Ch04

Chapter 4

The Energy of Particles 4.1

Kinetic Energy

75

4.1.1 GOAL: Find the work performed by the lift. GIVEN: m = 1100 kg, y¨ = 0.9 m/s2 , y2 − y1 = 2.0 m DRAW

FORMULATE EQUATIONS: The work performed from state 1 to state 2 is given by Z

y

W = y

2

N dy

(1)

1

The balance of forces on the car is N − mg = m¨ y ⇒ N = m(g + y¨) SOLVE: (2) → (1) ⇒

W = m(g + y¨)(y2 − y1 ) W = (1100 kg)(9.81 m/s2 + 0.9 m/s2 )(2.0 m) = 23, 562 N·m W = 23, 562 N·m

76

(2)

4.1.2 GOAL: Find the speed of the mass when it is 0.6 m from the wall GIVEN: m = 0.5 kg, k = 40 N/m, unstretched length=L0 = 0.3 m, x1 = 2 m, x2 = 0.6 m DRAW:

ASSUME: The surface is frictionless and the spring is linear. FORMULATE EQUATIONS: We’ll apply work/energy:





KE = KE + W 1

2

KE = 0 1

SOLVE: W

1−2



W



W

1−2

1−2

= −(40 N/m)[(

=

= −k

Z

2

1 x 2

2

Z

Ft dt =

x 2

x

1−2

−(k(x − L0 ))dx

1

(x − L0 )dx = −k[

1

x x2 − L0 x] 2 2 x 1

(0.6 m)2 (2 m)2 −(0.3 m)(0.6 m) )−( −(0.3 m)(2 m) )] = 56 J 2 2

KE = W 2

s

x





s

Z

⇒

1−2

KE = W

1−2

1 = 56 J = mv 2 2

⇒

v* = −14.97 * ı m/s

77

v = 14.97 m/s

4.1.3 GOAL: Find speed of arrow after it has moved 1.6 feet GIVEN: arrow’s weight and force profile DRAW:

FORMULATE EQUATIONS:



KE = KE + W1−2 1

2

SOLVE: m=

(20 oz) = 3.88 × 10−3 slug (16 oz/lb)(32.2 ft/s2 )

KE = 0 1

Z 1.6

W1−2

40e−3.2x dx]ft·lb

= [ 0

1.6 



40 = − e−3.2x 3.2

 ft·lb = 12.4 ft·lb 0

1 mx˙ 2 = W1−2 2 1 (3.88 × 10−3 slug)x˙ 2 = 12.4 ft·lb 2 x˙ = 80 ft/s

78

4.1.4 GOAL: Find the speed of a mass after being acted on by a given force. GIVEN: Initial speed, mass, force profile DRAW:

FORMULATE EQUATIONS:



KE = KE + W1−2 1

2

SOLVE: 1 1 KE = mv 2 = (50 kg)(3 m/s)2 = 225 N·m 2 2 1 2

Z

W1−2 = − [−100 − 50

e−1.1x ]dx (2 + x)

0

Evaluating with the MATLAB M-file quad yields W1−2 = −113 N·m 1 (50 kg)x˙ 2 = 225 N·m − 113 N·m 2 x˙ = 2.12 m/s

79

4.1.5 GOAL: Plot speed and acceleration of a cyclist going downhill GIVEN: Mass, slope, and drag forces DRAW:

*

*

b1 b2

*

* ı  cos θ sin θ − sin θ cos θ

FORMULATE EQUATIONS: *

*

*

−mg *  + F b 1 + N b 2 = −m¨ sb1

Force balance: *

*

*

*

*

−mg(sin θ b 1 + cos θ b 2 ) + (5 N + as˙ 2 ) b 1 + N b 2 = −m¨ sb1 *

m¨ s = mg sin θ − 5 N − as˙ 2

(1)

*

−mg cos θ + N = 0

(2)

b1 :

b2 :

Z

Work/displacement:

s2

Fds = W1−2

(3)

5 N 0.04s˙ 2 − m m

(4)

s1

SOLVE: s¨ = g sin θ −

(1) ⇒

Using (4) in MATLAB (θ = 6◦ , a = 0.04 N· s2 /m2 , m = 50 kg) produces the required s, s˙ data. Using this data along with (4) then lets us solve for s¨ and produce the following two plots.

80

To calculate the work done by drag forces we employ Wdrag = −

Z

50 m 



5 N + as˙ 2 ds

0

Using the speed/displacement data already obtained from the numerical simulation lets us calculate the work done and produces a final result of Wdrag = −340 N· m

81

4.1.6 GOAL: Determine the work needed to lift a block. GIVEN: Weight of block and change in height. DRAW:

FORMULATE EQUATIONS: The work done has a positive component (due to F ) and a R negative component (due to gravity). The total work done is found from W = 6 ft (F − mg)dx 1−2

and our energy equation is

0





KE + W1−2 = KE 1

(1)

2





SOLVE: The speed at state 1 and 2 is zero and therefore KE = KE = 0 1 2 Both forces acting on the block are constant and thus their work integral is simply the force times the displacement: (1) ⇒

F h − mgh = 0 ⇒ F h = mgh Work done by F = mgh = (50 lb)(6 ft) = 300 lb· ft

82

4.1.7 GOAL: Find required constant force to project rock to given height. GIVEN: Desired height and parameters. DRAW:

FORMULATE EQUATIONS: v32 − v22 = 2a(y3 − y2 )

Rectilinear motion with constant acceleration: Z

Work/energy:

y2

y1

SOLVE: (1) ⇒





F dy = W1−2 = KE − KE 2

1

(0 m/s)2 − v22 = −2g(10 m − 0.5 m) v22 = 2(9.81 m/s2 )(9.5 m)

(3) ⇒

(1) (2)

(3) (4)

0.5 Zm

1 (F − mg)dy = (F − (5 kg)(9.81 m/s2 ))(0.5 m) = (5 kg)2(9.81 m/s2 )(9.5 m) 2

(2) ⇒ 0

(5) ⇒

F = 981 N

83

(5)

4.1.8 GOAL: Determine the coefficient dynamic friction needed to bring a sliding mass to rest after rising 5 m. GIVEN: Mass, slope and initial speed. DRAW:

*

*

b1 b2

*

* ı  cos θ sin θ − sin θ cos θ

FORMULATE EQUATIONS: The FBD gives the forces on the mass as *

*

N b 2 − µN b 1 − mg *  In addition to this, we’ll use our work/energy formulation: 1 1 mv12 + W1−2 = mv22 2 2 SOLVE: The forces we need be concerned with (the ones that act along the path) are given by F = −µN − mg sin θ *

A force balance in the b 2 direction gives us N = mg cos θ and so our applied force becomes F = −mg(µ cos θ + sin θ) Applying work/energy from state 1 to state 2 gives us 1 mv12 − mg(µ cos θ + sin θ)d = 0 2 µ=

v12 − 2g sin θd 2g cos θd

=

(10.2 m/s)2 − 2(9.81 m/s2 )(0.5)(10 m) 2(9.81 m/s2 )(0.866)(10 m) µ = 0.035

84

4.1.9 GOAL: Determine the coefficient dynamic friction needed to bring a sliding mass to rest after rising 5 m. GIVEN: Mass, slope and initial speed. DRAW:

*

*

b1 b2

*

* ı  √ √ 1/ 2 1/ 2 √ √ −1/ 2 1/ 2

FORMULATE EQUATIONS: The FBD gives the forces on the mass as *

*

N b 2 + µN b 1 − mg *  In addition to this, we’ll use our work/energy formulation: 1 1 mv12 + W1−2 = mv22 2 2 SOLVE: The forces we need be concerned with (the ones that act along the path) are given by F = mg sin θ − µN √ * A force balance in the b 2 direction gives us N = mg/ 2 and so, using θ = 45◦ , our downslope force becomes mg(1 − µ) √ F = 2 First let’s consider the case of µ = 0. In this case we have mgd 1 0 + √ = mv22 2 2 v2 (9.81 m/s2 )(10 m) √ = 2 ⇒ v2 = 11.8 m/s 2 2 Next, we’ll let µ = 0.1: 0+

mgd(1 − µ) 1 √ = mv22 2 2

(9.81 m/s2 )(10 m)(1.0 − 0.1) 1 √ = v22 ⇒ v2 = 11.2 m/s 2 2 The mass is moving 0.604 m/s slower due to the friction, a 5.1% decrease.

85

4.1.10 GOAL: Determine the coefficient dynamic friction needed to bring a sliding mass to rest after rising 5 m. GIVEN: Mass, slope and initial speed. DRAW:

FORMULATE EQUATIONS: The FBD gives the forces on the mass as  (−mg + a + bx) * To solve the problem we’ll use the energy/work formula: 1 1 mv12 + W1−2 = mv22 2 2 SOLVE: First let’s find the contact speed. We’re given that the pinecone falls 30 m and so have 1 mgh = mv12 2 v1 =

p

2gh =

q

2(9.81 m/s2 )(30 m) = 24.3 m/s

The force along the path of travel is given by F = mg − a − bx and to determine the work done we’ll simply integrate with respect to displacement. Z d 1 2 (a + bx)dx = 0 mv1 + mgd − 2 0 x denotes distance traveled beneath the snow surface and d distance traveled when the pinecone finally comes to rest. 1 bd2 mv12 + mgd − ad − =0 2 2 bd2 1 + (a − mg)d − mv12 = 0 2 2 d2 + d2 +

mv12 2(a − mg) d− =0 b b

2[4 N − (0.3 kg)(9.81 m/s2 )] (0.3 kg)(24.3 m/s)2 d− =0 14 N/m 14 N/m d = 3.48 m

86

4.1.11 GOAL: Determine the launch speed of a payload from a catapult. GIVEN: Force acting along the direction of travel, size of the catapult and the initial and final angle of the catapult arm. DRAW:

*

e*r e*θ

* ı  − cos θ sin θ sin θ cos θ

FORMULATE EQUATIONS: The FBD gives the forces on the mass as Fθ e*θ + N e*r − mg *  To solve the problem we’ll use the energy/work formula: 1 1 mv12 + W1−2 = mv22 2 2 SOLVE: First we need the total force F that acts along the direction of travel: F = Fθ − mg cos θ Our work/energy expression becomes π 2

Z

0+ Fθ r

(10 N)(1.5 m)

π 4

1 (Fθ − mg cos θ)rdθ = mv22 2

π π π 1 − mgr(sin − sin ) = mv22 4 2 4 2

π π π 1 − (0.45 kg)(9.81 m/s2 )(1.5 m)(sin − sin ) = (0.45 kg)v22 4 2 4 2 v2 = 6.61 m/s

87

4.1.12 GOAL: Determine the average force acting on a bullet as it travels through a rifle barrel. GIVEN: Length of the barrel, mass of the bullet and exit speed. DRAW:

FORMULATE EQUATIONS: To solve the problem we’ll use the energy/work formula: 1 1 mv12 + W1−2 = mv22 2 2 SOLVE:   88 ft/s 700 mph = (700 mph) = 1027 ft/s 60 mph Our work/energy expression is given by 1 F ∆x = mv22 2 



1 2 oz    (1027 ft/s)2 F (4 ft) =  2 (16 oz/lb) 32.2 ft/s2 F = 511 lb

88

4.1.13 GOAL: Find the final speed of block B. GIVEN: µs = µd = 0.6, mA = 10 kg, mB = 20 kg, y2 − y1 = 0.5 m DRAW

FORMULATE EQUATIONS: The force balance on block A is *

mA x ¨ = T − FA

*

0 = N − mA g

ı :

 :

where the force of sliding friction is given by FA = µN . Thus, the force balance can be resolved into a single equation: mA x ¨ = T − µmA g The force balance on block B is mB y¨ = mB g − T Lastly, the energy balance on the system is









KEA + KEB = KEA + KEB + WA 2

2

1

1



1−2

+ WB

1−2

(1)

ASSUME: Let’s assume that the rope is inelastic, which yields the following constraints: y2 − y1 = x2 − x1 ⇒ y˙ = x˙ SOLVE: The work done on block A is Z



WA

1−2

x

= x

2

(T − µmA g)dx = (T − µmA g)(x2 − x1 )

1

The work done on block B is Z



WB

1−2

y

= y

2

(mB g − T )dx = (mB g − T )(y2 − y1 )

1

Using the constraint, the total work done on the system becomes

W

1−2

= (T − µmA g + mB g − T )(y2 − y1 ) = g(mB − µmA )(y2 − y1 ) 89

(2)

Recognizing that the kinetic energy of the system at State 1 is zero, (1) can be written as 1 1 m x˙ 2 + mB y˙ 2 = g(mB − µmA )(y2 − y1 ) 2 A 2 v u u y˙ = t2g

(2) → (3) ⇒ s

y˙ =

mB − µmA mA + mB

(3)

!

(y2 − y1 )

20 kg − (0.6)(10 kg) 2(9.81 m/s ) (0.5 m) = 2.14 m/s 10 kg + 20 kg 2





y˙ = 2.14 m/s

90

4.1.14 GOAL: Find the speed of A when the friction force is equal end opposite of the insertion force. GIVEN: m = 2 kg, F = 12 N, Fs = k(x0 + αx), k = 100 N/m, x0 = 0.05 m, α = 0.27, µ = 0.4, starts from rest at x = 0 DRAW:

ASSUME: The spring model only serves to model the force change during the insertion and isn’t viewed as itself requiring work to accomplish the compression. We only consider the work done by friction and by the 12 N insertion force. The spring/wall interface is frictionless. FORMULATE EQUATIONS: We’ll apply work/energy:





KE = KE + W 1

2

KE = 0 1

KE = W

⇒

2

1−2 x

Z 1−2

=

f

0

Fx dx

N*  − mg *  − Fs *  +F* ı − Ff * ı = m¨ x* ı

FBD=IRD SOLVE: *  :

N − mg − Fs = 0 ⇒

N = mg + Fs = mg + k(x0 + αx) 



Ff = µN = µ mg + k(x0 + αx)

In order to find W

1−2

we need to find xf . The goal is to find the speed of A when the friction

force is equal and opposite of the insertion force of 12 N. Find xf that satisfies this condition. 



F = µ mg + k(x0 + αxf )

xf =

( Fµ − mg) k

! 

− x0

1 α





  12 N − (2 kg)(9.81 m/s2 ) 1 =  0.4 − 0.05 m = 0.2 m 100 N/m 0.27 W

x

Z 1−2

=

f

0



Z

Fx dx =

0

x

f

F − Ff dx 

F − Ff = F − µ mg + k(x0 + αx) = A + Bx where A and B are chosen to simplify the expressions. h

i

A = F −µ(mg+kx0 ) = 12 N−(0.4) (2 kg)(9.81 m/s2 ) + (100 N/m)(0.05 m) = 2.15 N B = −µkα = −(0.4)(100 N/m)(0.27) = −10.8 N/m 91

W

W

Z 1−2

= 0

x

f

"

Bx2 1−2

f

= Axf +

2

= (2.15 N)(0.2 m) +



KE = W 2

1−2

(−10.8 N/m)(0.2 m)2 = 0.214 J 2

1 = 0.214 J = mv 2 2 v* = 0.463 * ı m/s

92

#

Bx2 xf (A + Bx)dx = Ax + 2 0

⇒

v = 0.463 m/s

4.1.15 GOAL: Find the force acting against a car as it brakes to a stop. GIVEN: Car weighs=2800 lb and goes from 60 mph to zero in 143 ft. DRAW:

FORMULATE EQUATIONS: We’ll apply work/energy:





KE = KE + W 2

1

1−2

X*

F = N*  − mg *  − Fb * ı

FBD: SOLVE:

m=

2800 lb = 87.0 slg 32.2 ft/s2

Applying work/energy gives us 1 0 = mv 2 + 2

143 Z ft

−Fb dx

0

Fb =

1 2 2 (87.0 slg)(88 ft/s)

143 ft Fb = 2.35×103 lb

93

4.1.16 GOAL: Determine the work done on an aluminum pellet as it travels through an imaginary gun’s barrel. Express the work in an equivalent manner to illustrate its size. GIVEN: Length of the barrel, mass of the pellet and exit speed. DRAW:

FORMULATE EQUATIONS: To solve the problem we’ll use the energy/work formula: 1 1 mv12 + W1−2 = mv22 2 2 SOLVE: The exit speed is given as one quarter of the speed of light. We’ll ignore relativistic effects in the following calculations. 3.00×108 m/s v1 = 0, v2 = = 7.49×107 m/s 4 Our work/energy expression is given by 1 W1−2 = mv22 2 1 W1−2 = (0.005 kg)(7.49×107 m/s)2 = 1.40×1013 N· m 2 Now let’s compare this to the work associated with lifting a 1500 kg car in a 1 g gravitational field. The work done will simply be equal to mgh where h is the height that the car is lifted. (1500 kg)(9.81 m/s2 )h = 1.40×1013 N· m h=

1.40×1013 N· m 8 2 = 9.54×10 m (1500 kg)(9.81 m/s )

Considering the fact that the distance from the earth to the moon is about 3.84×108 m, this means that the energy to shoot the gun is equal to that needed to move a normally sized car more than twice as far as from the earth to the moon. What does that tell you about the probability that someone could actually stand in one place while firing such a weapon?

94

4.1.17 GOAL: Determine the maximum impact speed of a vehicle such that it comes to rest just as it reaches the end of a deformable barrier. GIVEN: Force profile as a function of deflection and weight of the vehicle. DRAW:

FORMULATE EQUATIONS: To solve the problem we’ll use the energy/work formula: 1 1 mv12 + W1−2 = mv22 2 2 SOLVE: The force along the path of travel is given by 1

F = −a − bx 2 and to determine the work done we’ll simply integrate with respect to displacement. Z d 1 1 2 (a + bx 2 )dx = 0 mv1 − 2 0 x denotes distance traveled after contacting the barrier and d is the distance traveled when the car finally comes to rest. 1 2 3 mv12 − ad − bd 2 = 0 2 3

v=

v   u 2 3 u u 2 ad + bd 2 t 3

m

v   u 1 3 2 u 2 2 u 2 (5000 lb)(30 ft) + (3000 lb/ ft )(30 ft) u 3 v=u u 3800 lb t

32.2 ft/s2

v2 = 90.1 m/s

95

4.1.18 GOAL: Find compression of a forklift’s bumper. GIVEN: Bumper force characteristics, initial speed and forklift mass DRAW:

FORMULATE EQUATIONS:



KE = KE + W1−2 1

2

SOLVE: 1 1 KE = mv12 = (1000 kg)(1.5 m/s)2 = 1125 N·m 2 2 1

KE = 0 2

x

x

−(2.7×107 N/m2 )x3 2 = − (2.7×10 N/m )x dx = = −(9×106 N/m )x3 3 Z

W1−2

7

2

2

0

0

0 = 1125 N·m − (9×106 N/m)x3 ⇒ x = 0.050 m

96

4.1.19 GOAL: Find the speed of a block after a compressed spring has released GIVEN: Mass, spring constant, compression DRAW:

FORMULATE EQUATIONS: State 1 is with the spring fully compressed and mass stationary. State 2 is with the spring completely uncompressed. There are no nonconservative work terms. KE1 + P E1 = KE2 + P E2 SOLVE: 1 1 KE1 = 0; P ESP 1 = kx2 = (981 N/m) (0.08 m)2 = 3.14 N·m; P Eg1 = 0 2 2 1 KE2 = mx˙ 2 = (0.1 kg) x˙ 2 2 P ESP 2 = 0 



P Eg2 = mgh = (0.2 kg) 9.81 m/s2 (0.08 m) = 0.157 N·m 3.14 N·m = (0.1 kg) x˙ 2 + 0.157 N·m x˙ = 5.46 m/s

97

4.1.20 GOAL: Determine the tension in a pair of restraining strings and the height above the floor attained by the released block. GIVEN: Spring characteristics and mass of block. DRAW:

FORMULATE EQUATIONS: To solve the problem we’ll use the energy/work formula: 1 1 mv12 + W1−2 = mv22 2 2 SOLVE: We initially have a static force balance in the vertical direction: −2T cos 30◦ − mg + 2k∆x = 0 T =

2(500 N/m)(0.05 m) − mg = 50 N√− mg 2 cos 30◦ 3

We see from this that as the weight of the block (mg) increases, the tension decreases and, for a sufficiently heavy block, the strings would go slack. Now we move onto the case when the strings break. We’ll consider three total states. State 1 is with the block at its lowest point (0.01 m above the floor with the spring compressed 0.05 m). State 2 is when the spring has reached its fully extended state (block is 0.06 m above the floor with a speed to be determined). After state 2 the block is in a free trajectory and will reach a maximum height when gravity has decreased its speed to zero (state 3). State 1 to state 2: 2k(∆x)2 1 − mg(0.05 m) = mv22 (1) 2 2 From state 2 to state 3 we have 1 mv22 = mg∆y (2) 2 (1), (2) ⇒

2k(∆x)2 − mg(0.05 m) = mg∆y 2 ∆y = −0.05 m +

k(∆x)2 2mg

The total height h is the height at state 2 (0.06 m) plus the change in height from state 2 to state 3 (∆y). Thus we have 98

h = 0.06 m + ∆y = 0.01 m +

2k(∆x)2 (500 N)(0.05 m)2 = 0.01 m + 2mg 2m(9.81 m/s2 )

h = 0.01 m + 0.127mkg· m Clearly the mass of the block will alter the ultimate height attained (as expected from physical considerations). Note that the analysis assumes that the mass isn’t so great as to prevent the spring from completely extending. The way to determine if this assumption is valid is to simply evaluate h for a given value of m. If h exceeds 0.06 m, the unstretched spring length, then we know that m was “small enough” to match our assumption.

99

4.1.21 GOAL: Determine the speed of a mass particle at |θ| = 30◦ along a circular path. GIVEN: Particle’s mass, shape of path, initial velocity and position. DRAW:

FORMULATE EQUATIONS: To solve the problem we’ll use the energy/work formula: 1 1 mv12 + W1−2 = mv22 2 2 SOLVE: We’ll first determine if the mass has enough energy to reach the top of the hill with a finite speed. If so, we know that it will then move to the right side of the hill (θ negative) and thus will eventually reach θ = −30◦ . If it turns out not to have enough energy to reach θ = 0 then the conclusion is that it stops somewhere partway up, reverses direction and eventually reaches θ = 30◦ . The force due to gravity that acts against the mass along its trajectory is mg sin θ. From state 1 to state 2 we have Z 0 1 1 mg sin θrdθ = mv22 mv12 + 2 2 θ 0 v22 = 2



1 (1.25 m/s)2 − (9.81 m/s2 )(1 m)(1 − cos 20◦ ) 2



v22 = 0.379( m/s)2 v22 is positive, implying a real solution. Our conclusion is that it does reach θ = 0 and therefore will pass θ = −30◦ . Denoting its position at θ = 30◦ as state 3 we have Z −30◦ 1 1 2 mv1 + mg sin θrdθ = mv32 2 2 θ 0 v32

1 = 2 (1.25 m/s)2 − (9.81 m/s2 )(1 m)(cos(−30◦ ) − cos 20◦ ) 2





v*m = −1.73 e*θ m/s

100

4.1.22 GOAL: Determine the height that a sliding mass particle will obtain. GIVEN: Friction force, mass of particle and initial speed and position. DRAW:

FORMULATE EQUATIONS: To solve the problem we’ll use the energy/work formula: 1 1 mv12 + W1−2 = mv22 2 2 SOLVE: Z θ Z θ 1 0 0 mg sin θrdθ = 0 Ff rdθ − mv12 − 2 0 0 1 mv12 − Ff rθ0 + mgr(cos θ0 − 1) = 0 2 (0.05 kg)(1.2 m/s)2 −(0.6 N)(1.2 m)θ0 +(0.1 kg)(9.81 m/s2 )(1.2 m)(cos θ0 −1) = 0 θ0 = 0.0929 rad

101

4.1.23 GOAL: Determine the speed of a mass particle at |θ| = 30◦ along a circular path. GIVEN: Particle’s mass, shape of path, initial velocity and position. DRAW:

FORMULATE EQUATIONS: To solve the problem we’ll use the energy/work formula, applied twice. 1 1 mv12 + W1−2 = mv22 2 2 SOLVE: We’ll first determine what the speed of the mass is when it reaches state B, starting from state C. Next, we’ll consider the change in speed to to work done from state B to the final state A. From state C to state B we have Z 35◦ 1 1 2 2 mg sin θhdθ = mvB mvC + 2 2 0 Note that Ff = 0 in this phase. Using vC = 0 gives us 2 vB = 2 [−hg(cos 35◦ − 1)]

From geometry we have d = h/ tan 35◦ . Thus h = d tan 35◦ = 7 m and 2 vB = 24.85( m/s)2

Now we go from state B to A at a constant slope of 35◦ . The normal force N is seen to be equal to mg cos 35◦ and thus we know that a resisting force of µmg cos 35◦ acts to oppose m’s motion down the slope. 1 1 2 2 mvB + mg sin 35◦ d − µmg cos 35◦ d = mvA 2 2 2 vA = 24.85( m/s)2 + 2gd(sin 35◦ − µ cos 35◦ ) 2 vA = 24.85( m/s)2 + 80.39( m/s)2 = 105.2( m/s)2

v*A = 10.3 e*θ m/s

102

4.1.24 GOAL: Determine the speed of a skier at the end of a downhill run. GIVEN: Skier’s mass, shape of path, initial velocity and position. DRAW:

FORMULATE EQUATIONS: To solve the problem we’ll use the energy/work formula, applied twice, once during the straight frictional phase from A to B and then again along the curved, friction-free phase from B to C. 1 1 mv12 + W1−2 = mv22 2 2 SOLVE: We’ll first determine what the speed of the mass is when it reaches state B, starting from state A. Next, we’ll consider the change in speed to to work done from state B to the final state C. From state A to state B the frictional force opposing the acceleration due to gravity is equal to µd N = µd mg 2 . ! Z 60 m √ 3mg µd mg 1 2 0+ − ds = mvB 2 2 2 0 ! √ 3 0.08 1 2 2 − (9.81 m/s ) (60 m) = mvB 2 2 2 vB = 31.2 m/s Now we go from state B to C with a friction-free interface. We need to integrate along the circular arc, accounting for how the component of force due to gravity changes as the position along the arc changes. Z π rad 2 1 1 2 2 mvB + (100 m) mg cos θdθ = mvC π 2 2 rad 3 1 π π (32.2 m/s)2 + (100 m)(9.81 m/s2 ) sin rad − sin rad 2 2 3 



vC = 44.2 m/s

103







1 2 = vC 2

4.1.25 GOAL: Find the velocity of the mass when x = −5 m GIVEN: x˙ = −5.0 m/s, µ = 0.4, m = 3 kg DRAW:

FORMULATE EQUATIONS: *

*

N b 2 + S b 1 − mg *  = ma* *

a* = x ¨b1 *

*

*

*

*

N b 2 + S b 1 − mg(sin 20◦ b 1 + cos 20◦ b 2 ) = m¨ xb1 *

m¨ x = S − mg sin 20◦

(1)

*

0 = N − mg cos 20◦

(2)

N = mg cos 20◦

(3)

b 1: b 2:

(2)⇒

If the mass is slipping , then S = µN . Thus: m¨ x = µmg cos 20◦ − mg sin 20◦ = mg(µcos 20◦ − sin 20◦ ) The force acting on the mass in the x direction is mg(µcos 20◦ − sin 20◦ ). STATE 1: x = 0m STATE 2: x = −5 m



KE + W1−2 = KE 1

2

1 1 m(5.0 m/s)2 − mg(µcos 20◦ − sin 20◦ )(5 m) = KE = mv22 2 2 2

1 1 (5.0 m/s)2 − (9.81 m/s2 ) [(0.4)(0.9397) − 0.3420] (5 m) = v22 2 2 v2 = 4.66 m/s

104

4.1.26 GOAL: Analyze a particle in two different reference frames FORMULATE EQUATIONS: (a) YES Inertial reference frames can differ by a constant translational velocity. A stationary particle would appear to move (equal and opposite) when viewed from a moving frame. Thus the kinetic energy would differ. (b) YES For the same reason as for (a). The distance ”traveled” by the particle will vary along with its perceived velocity. (c) YES The changes in (a) and (b) are consistent and the correct work/energy relations will hold. Example: Frame 1 is stationary. Frame 2 moves to the left at v0 m/s. A constant force F pushes a mass particle m to the right for t seconds. Frame 1: * * F * Since F = ma, we have a* = m . t2 F t2 ∆x = a = 2 2m W1−2 = F ∆x =

F 2 t2 2m





KE + W1−2 = KE 1

0+

F 2 t2 2m

= KE

2

2

F 2 t2 KE =

⇒

2

2m

(1)

Check: v after t seconds is given by: v = at =

Ft m

1 1 F 2 t2 F 2 t2 KE = mv 2 = m 2 = 2 2 m 2m 2

(2)

(1) and (2) match, as expected. Now look at the system from frame 2: Initially the mass is stationary. Since the frame is moving left at v0 , the perceived velocity is v0 (to the right). 1 KE = mv02 2 1 In t seconds the mass is perceived to move its actual change in displacement plus the amount the frame has moved: F t2 t2 ∆x = a + v0 t = + v0 t 2 2m W1−2 = F ∆x =

105

F 2 t2 + F v0 t 2m

1 F 2 t2 KE = KE + W1−2 = mv02 + + F v0 t 2 2m 1 2

(3)

Check: v after t seconds is given by: v = at + v0 =  2 1 Ft 1 + v0 = KE = m 2

2

m

KE = 2

(3) and (4) agree, showing that work/energy holds. (d) FALSE Work/energy was derived from Newton’s second law.

106

2

Ft + v0 m F 2 t2 2v0 F t + + v02 m2 m

F 2 t2 1 + F v0 t + mv02 2m 2

!

(4)

4.2

Potential Energies and Conservative Forces

107

4.2.1 GOAL: Find v when compression is 0.8 m. Discuss. GIVEN: System parameters and target compression amount. DRAW:

FORMULATE EQUATIONS: We’ll apply conservation of energy







KE + PE = KE + PE 1

1

2

2

SOLVE: 1 1 0 + mgh = mv 2 + kx2 2 2 1 1 (5 kg)(9.81 m/s2 )(1 m) = (5 kg)v 2 + (400 N/m)(0.8 m)2 2 2 v 2 = −31.6(m/s)2 ⇒ v is imaginary b) Imaginary velocities aren’t physically possible. Thus mathematics are telling us that the spring can’t compress .8 m. The maximal compression must be less than 0.8 m. Check: Find x when v = 0 1 mgh = kx2 2 1 (5 kg)(9.81 m/s2 )(1 m) = (400 N/m)x2 ⇒ x = 0.5 m < 0.8 m! 2

108

4.2.2 GOAL: Find θ for loss of contact. DRAW:

*

*

er e*θ

* ı  − sin θ − cos θ − cos θ sin θ

FORMULATE EQUATIONS: 2 −mg *  − N e*r = m[−rθ˙ e*r + rθ¨e*θ ]

Force balance:

Our second equation involving θ˙ and θ comes from conservation of energy: KE + PE = KE + PE 1

1

2

2

(1)

(2)

SOLVE: For loss of contact we’d want N = 0 and thus 2

rθ˙ = −g cos θ (2) ⇒ (3) → (4) ⇒

1 1 ˙ 2 + mgr(1 − cos θ) mv12 = m(rθ) 2 2 1 1 mv12 = − mrg cos θ + mgr(1 − cos θ) 2 2 3 1 mv12 = mgr(1 − cos θ) 2 2 1 3 (4 kg)(7.95 m/s)2 = (4 kg)(9.81 m/s2 )(1.5 m)(1 − cos θ) 2 2 cos θ = −0.765 θ = 140◦

109

(3) (4) (5) (6)

4.2.3 GOAL: Find the distance travelled by the mass. GIVEN: k = 100 N/m, mA = 0.05 kg, µs = 0.6, µd = 0.3, initial spring compression of 0.1 m DRAW

*

* ı  ◦ cos 20 sin 20◦ ◦ − sin 20 cos 20◦

*

b1 b2

*

FORMULATE EQUATIONS: Let’s first check if the force of static friction is sufficient to hold the mass against the compressed spring. *

where Ff

max

*

*

mA s¨b 1 = Fspring b 1 − Ff

Force balance:

max

*

b 1 + N b 2 − mA g * 

= µs N , Fspring = kx, and x is the compression of the spring. *

SOLVE: Looking just at the b 1 component of the force balance gives us mA s¨ = kx − mA g(µ cos 20◦ + sin 20◦ ) mA s¨ = (100 N/m)(0.1 m) − (0.05 kg)(9.81 m/s2 )(0.6 cos 20◦ + sin 20◦ ) = 9.56 N *

Thus, there is a positive force in the b 1 direction even when the force of static friction is at its maximum. So we can now solve the problem knowing that dynamic friction is acting on the mass. FORMULATE EQUATIONS: Let’s now look at the case in which the mass has started at rest while pushed up against the spring, and ends up a distance d further up the ramp. Because the mass starts at rest, and ends at rest, the energy balance is





PE = PE + Wnc 1

2

1−2

SOLVE: The non-conservative work done by friction is:

Wnc

1−2

s

Z



Wnc

1−2

= s

2

s

Z

Ff ds =

1

s

2

(−µd mA g cos 20◦ )ds

1

= −µd mA g(s2 − s1 ) cos 20◦ = −µd mA gd cos 20◦

The expressions for potential energy at the two states are 1 PE = mA gs1 sin 20◦ + kx2 2 1

PE = mB gs2 sin 20◦ 2

Plugging these into (1) gives us 1 mA gd sin 20◦ = kx2 − µd mA gd cos 20◦ 2 110

(1)

Solving for d yields d=

1 2 2 kx mA g(sin 20◦ + µd

cos 20◦ )

=

0.5(100 N/m)(0.1 m)2 = 1.63 m (0.05 kg)(9.81 m/s2 ) [sin 20◦ + (0.3) cos 20◦ ] d = 1.63 m

111

4.2.4 GOAL: Find the cushion’s maximum compression. GIVEN: m = 0.2 kg, h = 1.5 m, k = 40 N/m DRAW:

ASSUME: The track is frictionless. FORMULATE EQUATIONS: We’ll apply conservation of energy:







KE + PE = KE + PE 2

2

1

PE = PE 2

1

1

SOLVE: 0.5kx2 = mgh s

x=

2mgh = k

s

2(0.2 kg)(9.81 m/s2 )(1.5 m) = 0.384 m (40 N/m) x = 0.384 m

112

4.2.5 GOAL: Two methods to impart velocity to a mass are proposed. Evaluate them and determine if one will produce a higher velocity. GIVEN: Starting and ending configuration of the two methods. DRAW:

FORMULATE EQUATIONS: All we need is to apply conservation of energy:







KE + PE = KE + PE 1

1

2

2

SOLVE: Case A: 1 my˙ 2 + 0 2 p y˙ = 2gh √ v* impact = − 2gh * 

0 + mgh =

Case B: ˙ The speed at impact will be given by hθ. 1  ˙ 2 m hθ + 0 2 p hθ˙ = 2gh

0 + mgh =

√ v* impact = −hθ˙ * ı = − 2gh * ı CHECK: The impact velocities are identical and therefore both methods will crack coconuts equally well. From a design perspective I would have to say Case B would make for more repeatable and accurate strikes but Case A has the advantage of no moving parts so I would have to call it a tie. 113

4.2.6 GOAL: Find spring constant to limit spring compression to specified amount. GIVEN: Initial speed and parameters. L = 5 m, m = 2 kg, θ = 30◦ , v = 8 m/s. DRAW:

FORMULATE EQUATIONS: We’ll apply conservation of energy: KE + PE = KE + PE 2

SOLVE:

1

1

1 1 0 + k(0.1 m)2 + mg(L + 0.1 m) sin θ = mv 2 2 2

(1) ⇒

(2) ⇒

2

k=

(2 kg)((8 m/s)2 − (9.81 m/s2 )(5.1 m)) = 2.79×103 N/m 0.1 m2

114

(1)

(2)

4.2.7 GOAL: Determine the impact speed of a falling cyclist. GIVEN: Cyclist’s initial orientation and distance from the ground. DRAW:

FORMULATE EQUATIONS: We will employ conservation of energy from state 1 to state 2: KE + PE = KE + PE 1

1

2

2

SOLVE: We’re told that the cyclist simply topples over and thus at the point of impact with the ground is moving straight down (from geometry). Our energy conservation equation becomes 1 0 + mgh = mv 2 + 0 2 The mass drops out and we’re left with 1 (9.81 m/s2 )(1 m) = v 2 2 v = 4.43 m/s

115

4.2.8 GOAL: Determine the impact speed of a mass against the ceiling GIVEN: Initial configuration of the system and spring constants. DRAW:

FORMULATE EQUATIONS: We will employ conservation of energy from State 1 to State 2: KE + PE = KE + PE 1

1

2

2

SOLVE: √ Initially (State 1) the kinetic energy is zero and the springs are stretched by an amount 42 + 22 m− 1 m and at State 2 are stretched (2 m − 1 m). Our energy conservation equation is 2 1 1 1 p 2 4 + 22 m − 1 m = mv 2 + 2 k(2 m − 1 m)2 + mg(4 m) 2 k 2 2 2 1 1 1 2 (80 N/m) (3.472 m)2 = (10 kg)v 2 +2 (80 N/m)(1 m)2 +(10 kg)(9.81 m/s2 )(4 m) 2 2 2 v = 9.92 m/s

116

4.2.9 GOAL: Determine the maximum compression of a spring that models the elastic elements of a leg. GIVEN: Initial configuration of the system, mass and spring constant. DRAW:

FORMULATE EQUATIONS: We will employ conservation of energy from state 1 to state 2: KE + PE = KE + PE 1

1

2

2

SOLVE: At State 1 the spring has not yet compressed. At State 2 the speed is zero. Our conservation of energy equation is 1 1 mv 2 = kh2 − mgh 2 2 where h is the compression of the leg spring. 1 1 (10 kg)(2 m/s)2 = (14, 000 N/m)h2 − (10 kg)(9.81 m/s2 )h 2 2 h = 6.09 cm

117

4.2.10 GOAL: Determine the maximum height off the ground a rebounding toy will reach. GIVEN: Initial configuration of the system, mass and spring constant. DRAW:

FORMULATE EQUATIONS: We will employ conservation of energy from State 1 to State 2: KE + PE = KE + PE 1

1

2

2

SOLVE: At State 1 the spring is fully compressed and the mass has zero speed. At State 2 the speed is again zero. Our conservation of energy equation is 1 k(0.04 m − 0.008 m)2 = mg(h − 0.008 m) 2 1 (3000 N/m)(0.04 m − 0.008 m)2 = (0.08 kg)(9.81 m/s2 )(h − 0.008 m) 2 h = 1.97 m

118

4.2.11 GOAL: Determine the effective spring constant of a trampoline. GIVEN: Maximum deflection of the trampoline as a result of a falling body and the mass of the body. DRAW:

FORMULATE EQUATIONS: We will employ conservation of energy from State 1 to State 2, State 2 to 3, and State 3 to 4. SOLVE: We’re given that the upward speed at State 1 is 12 ft/s. After reaching a maximum height, the mass will come back down and, arriving at the same height at State 2, will have the same speed but oriented down instead of up. Thus v2 = v1 = 12 ft/s. We can apply conservation of energy from State 2 to 3 and then from 3 to 4 in a single step. We’ll define the zero potential energy state as that associated with the mass at the level of the undeformed trampoline, as in State 3. 1 1 m(12 ft/s)2 + mg(50 ft) = −mg(1.3 m) + k(1.3 ft)2 2 2 1 2

150 lb 32.2 ft/s2

!

1 (12 ft/s)2 + (150 lb)(50 ft) = −(150 lb)(1.3 m) + k(1.3 ft)2 2 k = 9503 lb/ft

119

4.2.12 GOAL: Find the speed with which a mass particle hits an elastically mounted stop after rebounding once. GIVEN: System parameter values. DRAW:

*

*

b1 b2

*

* ı  cos θ sin θ − sin θ cos θ

FORMULATE EQUATIONS: We’ll use work/energy







KE + PE + W1−2 = KE + PE 1

1

2

2

and break the analysis into segments. SOLVE: Initially we have a potential energy of mg(5 m) and zero kinetic energy. Negative work is done as the mass drops due to friction, with the work equal to −µN d = −µmgd cos θ (where N = mg cos θ is the normal force between the mass and slope and d is the distance moved along the slope). At the end of this first phase of motion the mass has dropped a distance h below it’s zero point (the vertical position of the undeformed spring pad. Our energy balance is therefore     5m + h 1 h 2 0 + mg(5 m) − µmg cos θ = −mgh + k sin θ 2 sin θ Using θ = 40◦ and solving with the given parameter values (realizing that k = 150, 000 N/m) yields h = 1.015×10−2 m. This is quite a small deflection and reflects the fact that the spring basically acts to rebound the mass without much “give” of its own. Thus at State 2 we have zero speed, the mass has moved slightly below its original zero potential energy position and we have a negative gravitational potential energy and a positive spring potential energy. Next we calculate how high the mass will reach upon rebound:     h + y3 1 h 2 k − mgh − µmg = mgy3 2 sin θ tan θ Solving this for the given parameter values and using h = 1.015×10−2 m yields a rebound height of y3 = 3.071 m. Our final calculation equates the potential energy at a height y3 minus the loss due to the final slide down the incline with the kinetic energy at the spring pad:  y3 1 mgy3 − mgµ = mv 2 tan θ 2 v = 6.94 m/s 120

4.2.13 GOAL: Find the speed with which hinge B is moving after the system has moved under the influence of a stretched spring. GIVEN: Initial and final system configuration, link lengths, masses and spring constant. DRAW:

*

*

b1 b2

*

* ı  cos θ sin θ − sin θ cos θ

FORMULATE EQUATIONS: We’ll use work/energy for this no-loss system







KE + PE = KE + PE 1

1

2

2

The challenge here is to correctly determine the speed of both B and C, which we’ll do by use of the law of sines: x BC = sin γ sin θ where the angles and sides are illustrated in State 2 of the figure. SOLVE: At State 1 we have a total energy of 1 1 E = mB gh+ k(1.2 m−0.2 m)2 = (10 kg)(9.81 m/s2 )(0.4 m)+ (80 N/m)(1 m)2 = 79.24 N·m (1) 2 2 At State 2, C is moving to the right at a speed x˙ and B is rotating about A with a speed |vB | = ˙ (0.5 m)|θ|. BC x = ⇒ x sin θ = BC sin γ = BC sin(180◦ − θ − β) sin γ sin θ Differentiating with respect to t gives us ˙ x˙ sin θ + xθ˙ cos θ = −BC cos(180 − θ − β)(θ˙ + β)

(2)

˙ Looking at State 2 in the figure gives us (from To go further we need to find β˙ in terms of θ. geometry) (0.5 m) sin θ = (0.4 m) sin β and 121

(3)

x = (0.5 m) cos θ + (0.4 m) cos β

(4)

Differentiating (3) yields (0.4 m)β˙ cos β = (0.5 m)θ˙ cos θ β˙ =



(0.5 m) cos θ (0.4 m) cos β



θ˙

(5)

Using (3) with θ = 30◦ gives us β = 38.68◦ and (4) yields x = 0.745 m. Using these values of x and β, along with (5), and substituting them into (2) gives us θ˙ = −(1.676 m−1 )x˙ We can now evaluate the energy at State 2. C has moved to the right and the total stretch of the spring is now (1.2 m − (0.745 m − 0.3 m) − 0.2 m) = 0.555 m. The energy when θ = 30◦ is 1 1 1 2 2 E = k(0.555 m)2 + mB g(0.25 m) + mC vC + mB vB 2 2 2 1 1 1 ˙2 E = (80 N/m)(0.555 m)2 +(10 kg)(9.81 m/s2 )(0.25 m)+ (15 kg)x˙ 2 + (10 kg)[(0.5 m)(−1.676 m−1 )x] 2 2 2 E = 36.83 N· m + (11.01 kg)x˙ 2 (1), (6) ⇒

x˙ = 1.96 m/s ˙ = |(0.5 m)(−1.676 m−1 )(1.96 m/s)| |vB | = |(0.5 m)θ| |vB | = 1.64 m/s

122

(6)

4.2.14 GOAL: Find if collar reaches B and, if so, its speed at B. GIVEN: Mass of collar, spring constant, system geometry and initial speed DRAW:

FORMULATE EQUATIONS:



KE + PE 1



g 1

+ PE



sp



= KE + PE 2

1



g 2

+ PE

sp

2

SOLVE: 1 KE = mv12 = 0 2 1

PE

PE

sp

1

g 1

=0

1 1 = kx21 = (30 N/m)(0.7 m − 0.1 m)2 = 5.4 N·m 2 2 1 KE = mv22 = (0.25 kg)v22 2 2



PE

g 2



PE

sp

2

= mgh = (0.5 kg)(9.81 m/s2 )(0.4 m) = 1.962 N·m 1 1 = kx22 = (30 N/m)(0.4 m − 0.1 m)2 = 1.35 N·m 2 2 (0.25 kg)v22 = (5.4 − 1.962 − 1.35)N·m v2 = 2.89 m/s

Clearly, the mass reaches B, as shown by the real solution for v2 .

123

4.2.15 GOAL: Determine the minimum spring constant k so that a mass does not impact the bottom of a dropped enclosure. GIVEN: Size of enclosure, number of springs and arrangement, mass and clearance in enclosure. DRAW:

FORMULATE EQUATIONS: We will employ conservation of energy from State 1 to State 2: PE + KE = PE + KE 1

1

2

2

SOLVE: We’re given that the speed of the enclosure and mass just before impact is 8 m/s. Thus the mass will still have this speed even though the enclosure is brought to an abrupt stop. At State 1 the mass has a finite kinetic energy and also a finite potential energy due to its height above the ground. At State 2 we assume zero speed, zero potential energy due to gravity and a potential energy due to the extension/compression of the positioning springs.







PE + KE = PE + KE 1

1

2

2

1 1 mgh + mv 2 = (2k)h2 2 2 1 1 (0.02 kg)(9.81 m/s2 )(0.04 m) + (0.02 kg)(8 m/s)2 = (2k)(0.04 m)2 2 2 k = 404.9 N/m

124

4.2.16 GOAL: Find vB after |θ| = 45◦ GIVEN: geometry and masses DRAW:

FORMULATE EQUATIONS: To determine rotation, perform a moment balance about O X

MO = mB gr − mA gL sin 30◦

(1)

followed by conservation of energy:







KE + PE = KE + PE 2

1

1

(2)

2

SOLVE: 



(1) ⇒ mB gr − mA gL sin 30◦ = 9.81 m/s2 [(0.5 kg) (0.08 m) − (1.1 kg) (0.6 m) (0.5)] = −2.84 N·m The moment sum is negative and thus the disk rotates clockwise

KE = 0 1

  PE g = mA gL cos 30◦ = (1.1 kg) 9.81 m/s2 (0.6 m) (0.866) = 5.61 N·m 1

1 1 KE = L2 θ˙2 mA + r2 θ˙2 mB 2 2 2 



PE g = mB gr 2





= (0.5 kg) 9.81 m/s2 (0.08 m)



π π rad − rad + mA gL cos 45◦ 4 6 

  π π 1 rad − rad + (1.1 kg) 9.81 m/s2 (0.6 m) √ 4 6 2 





= 4.68 N·m (2) ⇒

i 1 h 5.61 N·m = 4.68 N·m + θ˙2 (0.6 m)2 (1.1 kg) + (0.08 m)2 (0.5 kg) 2

θ˙ = 2.15 rad/s 125

v*B = rθ˙ *  = (0.08 m) (2.15 rad/s) *  = 0.172 *  m/s

126

4.2.17 GOAL: Determine the speed of a mass (part of a two-mass/spring system) when the orientation of the supporting link has moved from its initial position to a new one. GIVEN: Inclination angle is initially 60◦ and finally 45◦ . Mass and spring constants as well as system geometry. DRAW:

FORMULATE EQUATIONS: We will employ conservation of energy from State 1 to State 2: KE + PE = KE + PE 2

1

1

2

SOLVE: Initially (State 1) the kinetic energy is zero and the spring is stretched by an amount 2(1.2 m) sin 60◦ − 1.23 m = 0.848 m. At State 2 the spring is stretched by 2(1.2 m) sin 45◦ − 1.23 m = 0.467 m. Both masses will be moving at the same speed (from symmetry) and we’ll call this speed v. Our energy conservation equation is KE + PE = KE + PE



1



1



2

2

1 1 1 k (0.848 m)2 +2mg(1.2 m) cos 60◦ = (2m)v 2 + k (0.467 m)2 +2mg(1.2 m) cos 45◦ 2 2 2 1 (42 N/m) (0.848 m)2 + 2(2.0 kg)(9.81 m/s2 )(1.2 m) cos 60◦ = 2 1 1 (2)(2.0 kg)v 2 + (42 N/m) (0.467 m)2 + 2(2.0 kg)(9.81 m/s2 )(1.2 m) cos 45◦ 2 2 v = 0.626 m/s From geometry we see that B’s velocity is oriented up and to the left: ı + v*B = 0.626(− √12 *

127

√1 * ı ) m/s 2

= 0.443(− * ı +*  ) m/s

4.2.18 GOAL: Determine the speed of a mass (part of a mass/two spring system) when it has dropped a set distance beneath its release position. GIVEN: Mass and spring constants, initial and final system orientation. DRAW:

FORMULATE EQUATIONS: We will employ conservation of energy from State 1 to State 2: KE + PE = KE + PE 2

1

1

2

SOLVE: Initially (State 1) √ the kinetic energy is zero and the springs are unstretched. At State 2 each spring is stretched by 0.012 + 0.042 m − 0.01 m = 0.03123 m. The mass will be moving at speed v. Our energy conservation equation is KE + PE = KE + PE 1









1

2

2

1 1 mg(0.04 m) = mv 2 + 2 k (0.03123 m)2 2 2  

1 1 (60 N/m) (0.03123 m)2 (0.5 kg)(9.81 m/s2 )(0.04 m) = (0.5 kg)v 2 + 2 2 2  

v = 0.742 m/s We know the mass is dropping down and thus have v*B = −0.742 *  m/s

128

4.2.19 GOAL: Find the distance the cyclist can coast. GIVEN: v1 = 25 mph, 6% grade, no air resistance or friction DRAW

*

*

b1 b2

*

* ı  cos θ sin θ − sin θ cos θ

FORMULATE EQUATIONS: The energy balance is given by







KE + PE = KE + PE 1

2

2

(1)

1

SOLVE:







PE − PE = KE − KE

Rearranging (1) ⇒

2

1

1



2

(2)

Because the cyclist will be at rest at state 2, KE = 0. The change in potential energy in terms of 2 d is given by PE − PE = mgd sin θ (3) 2

1

1 mgd sin θ = mv12 2

(3) → (2) ⇒ 

d=

v12 2g sin θ

=

ft/s 25 mph × 1.4667 mph

2

2(32.2 ft/s2 ) sin [tan−1 (0.06)] d = 348.6 ft

129

= 348.6 ft

4.2.20 GOAL: Find the speed at which the separated block hits the ceiling. GIVEN: m = 10 kg, k = 200 N/m DRAW:

ASSUME: The spring has an unstretched length of zero and will not provide a force when the block hits the ceiling. FORMULATE EQUATIONS: We’ll apply conservation of energy: KE + PE = KE + PE 1

1

2

2

KE + PE = PE

1

KE = PE − PE

2

2

2

2

1

SOLVE: First use equilibrium to find yeq , the initial stretched length of the spring. kyeq *  − mg *  =0 ⇒

FBD=0

yeq =

mg k

1 k mg 2 m2 g 2 PE = k(yeq )2 = ( ) = 2 2 k 2k 1

PE = 2

m mg mg mg m2 g 2 gh = (yeq ) = ( )= 2 2 2 k 2k







KE = PE − PE = 2

1

2

KE = 0 2

⇒

v=0

130

m2 g 2 m2 g 2 − =0 2k 2k v=0

4.2.21 GOAL: Find speed when fishing pole is straight. GIVEN: Initial conditions and parameters. DRAW:

FORMULATE EQUATIONS: We’ll apply conservation of energy: KE + PE = KE + PE 1

2

2

1

(1)

An equilibrium force balance before the fish drops off gives us ky − (mA + mB )g = 0 where y is the distance that the end of the rod has sagged under a gravitational load. SOLVE: (mA + mB )g k= (2) ⇒ y (1), (3) ⇒

(3), (4) ⇒

1 1 mA v 2 = ky 2 − mA gy 2 2

(2)

(3) (4)

1 1 (0.04 kg + 10 kg)(9.81 m/s2 ) (0.04 kg)v 2 = (0.2 m)2 − (0.04 kg)(9.81 m/s2 )(0.2 m) 2 2 0.2 m v = 22.1 m/s

131

4.2.22 GOAL: Determine the variation in speed of a sliding block under friction-free vs friction conditions. GIVEN: Angle of sloped surface, coefficient of friction. The mass slides 2 m along the slope. DRAW:

*

*

b1 b2

*

* ı  ◦ cos 15 sin 15◦ − sin 15◦ cos 15◦

FORMULATE EQUATIONS: In both cases we’ll utilize work/energy including both potential and kinetic energy terms: KE + PE + W1−2 = KE + PE



1



1



2

2

(1)

SOLVE:

First we’ll look at the zero friction case. Using (1) with no included dissipative work term, along with a height change of h1 , gives us







KE + PE = KE + PE 1

1

2

2

1 mgh1 = (5 kg)s˙ 2 2 1 (5 kg)(9.81 m/s2 )(2 sin 15◦ m) = (5 kg)s˙ 2 2 s˙ = ±3.19 m/s We know from the geometry of the problem that the mass is moving downslope when it reaches State 1 and so we have * v*B = −3.19 b 1 m/s

132

Now consider a coefficient of friction µ = 0.2.

*

*

*

Force balance:

m¨ s b 1 = F b 1 + N b 2 − mg * 

Force balance:

m¨ s b 1 = F b 1 + N b 2 − mg(sin 15◦ b 1 + cos 15◦ b 2 )

*

*

*

*

*

*

*

*

m¨ s b 1 = F b 1 − mg sin 15◦

(2)

*

0 = N − mg cos 15◦ ⇒ N = (5 kg)(9.81 m/s2 ) cos 15◦ = 47.38 N

(3)

b1 : b2 :

First let’s verify that the mass does, indeed, slip downslope. If we assume that s¨ = 0 then (2) implies F = mg sin 15◦ = 12.7 N Thus, the force needed to keep the mass from accelerating downslope is 12.7 N. The maximum frictional force obtainable is given by Fmax = µN = 0.2(47.38 N) = 9.48 N The maximum frictional force is less than that needed to hold the mass stationary and therefore we have slip. KE + PE + W = KE + PE (1) ⇒ 1−2 1 1 2 2 1 mgh1 − |F |(2 m) = ms˙ 2 2 1 (5 kg)(9.81 m/s2 )(2 m sin 15◦ ) − (9.48 N)(2 m) = (5 kg)s˙ 2 2 s˙ = ±1.60 m/s Again, we know the mass moves downslope and therefore have *

v*B = −1.60 b 1 m/s The addition of the friction slowed the mass by 100



3.19−1.60 3.19



= 0.50, i.e. by 50 percent

133

4.2.23 GOAL: Determine if a trampoline bottoms out when a person jumps into it and, if so, how high the supports must be raised to avoid bottoming. GIVEN: Height that person jumps from (20 ft), spring constant of trampoline (40 lb/in) and person’s weight (122 lb). DRAW:

FORMULATE EQUATIONS: m=

122 lb = 3.79 slg, k = (40 lb/in)(12 in/lb) = 480 lb/ft 32.2 ft/s2

We’ll approach the problem in two steps. First we’ll find the speed of impact with the undeformed trampoline (a fall of 17 ft). Then we’ll assume a zero velocity state (State 3) associated with a deflection h of the trampoline. If h is less than 3 ft then we can conclude the trampoline doesn’t bottom out. If h is more than 3 ft then we immediately have the required height. SOLVE: Assume that when at State 2 the falling mass has zero velocity. Because we start with zero velocity as well, there is no kinetic energy term in our energy balance. PE + KE = PE + KE 2

1

1









2

1 mg(17 ft) = mv 2 2 1 (122 lb)(17 ft) = (3.79 slg)v 2 2 v 2 = 1095( ft/s)2 Now we proceed from State 2 to State 3:







PE + KE = PE + KE 2

2

3

3

1 1 mgh + mv 2 = kh2 2 2 1 1 (3.79 slg)(1095( ft/s)2 ) + (3.79 slg)(32.2 ft/s2 )h = (4880 lb/ft)h2 2 2 h2 − (0.508 ft)h − 8.65 ft2 = 0 h = 3.19 ft Since 3.19 ft > 3 ft we see that the answer to part (a) is yes - the trampoline will bottom out. The answer to (b) is that to avoid bottoming out, the trampoline must be at least 3.19 ft above the ground.

134

4.2.24 GOAL: Verify the numerical results from Problem 3.5.28 using energy methods. Find the coefficient of friction which causes the sliding mass to stop at θ = 30◦ . GIVEN: r = 20 m, θ˙ = 0.5 rad/s and µd = 0.1. The following is a program to calculate the kinetic energy of the mass particle as work is done to it from friction and gravity. The plots show the exact kinetic energy ( 21 mv 2 ) versus the kinetic energy calculated from a work standpoint. As you can see, the results are in agreement.

135

136

4.2.25 GOAL: Determine the velocity of a mass particle after traveling through a tube. GIVEN: The size and shape of the tube, mass of the particle and initial velocity. DRAW:

FORMULATE EQUATIONS: We’ll use both energy conservation







KE + PE = KE + PE 1

1

2

2

SOLVE: Because the tube’s surface is frictionless, it doesn’t matter that the mass is traveling along something other than a straight, vertical path - the kinetic energy change is simple due to the change in potential energy from the start to the finish. All forces generated between the mass and the walls of the tube are normal to the direction of travel and hence do no work. KE + PE = KE + PE 1

1

2

2

1 1 mv12 = mv22 + mg(2 m) 2 2 1 1 (0.1 kg)(11 m/s)2 = (0.1 kg)v22 + (0.1 kg)(9.81 m/s2 )(2 m) 2 2 v*2 = 9.04 *  m/s

137

4.2.26 GOAL: Determine if an actor can successfully swing across a chasm. GIVEN: The breaking strength of the vine is 1100 N and the actor has a mass of 80 kg. The total includes angle of the vine from start to finish is 60◦ . DRAW:

FORMULATE EQUATIONS: We’ll use both energy conservation







KE + PE = KE + PE 2

1

1

2

and a force balance at the bottom of the swing (T − mg) *  = m(rθ˙2 *  + rθ¨* ı) SOLVE: * ı :

0 = mrθ¨ T − mg = mrθ˙2 ⇒ T = mg + mrθ˙2

*

 :

(1)

Now let’s apply conservation of energy.







KE + PE = KE + PE 1

1

2

2

1 1 mv12 + mgh1 = mv22 + mgh2 2 2 Using h1 = (10 m)(1 − cos 30◦ ) = 1.34 m, and h2 = 0 we have 1 (80 kg)(9.81 m/s2 )(1.34 m) = (80 kg)v22 ⇒ |v2 | = 2 v2 = 5.127 m/s ˙ 2 ⇒ θ˙2 = 0.2629 ( rad/s)2 v22 = [(10 m)θ] (1), (2) ⇒

(2)

T = (80 kg)(9.81 m/s2 ) + (80 kg)(10 m)(0.2629( rad/s)2 ) = 995 N

Because the vine’s tensile strength is 1,100 N, more than the needed 995 N, the stunt can be successfully completed.

138

4.2.27 GOAL: Find the velocity of Block A when mass B strikes the ground. GIVEN: Bar is massless and has length L, torsional spring applies M = −kθ θ mA = 5 kg, mB = 10 kg, L = 0.8 m, kθ = 10 kg·m/rad DRAW:

FORMULATE EQUATIONS: State 1: Pendulum upright State 1: Pendulum horizontal

KE = 0

(1)

1

1 PE = mB gL + kθ 2 1

 2

π 2

1 1 2 2 KE = mA vA + mB vB 2 2 2

(2) (3)

PE = mA g(sin(30◦ ))∆xA

(4)

2∆xA = ∆y

(5)

2

where ∆xA is the motion of block A up the slope and ∆y is the distance the rope going over the pulley moves. The figure entitled p Labeling shows that the length of rope from the pulley C to mass B is initially 0.5L and finally (1L)2 + (1.5L)2 = 1.8L.

∆length of rope = 1.80L − 0.5L = 1.30L Therefore: 139

∆y = 1.30L = 2∆xA ∆xA = 0.65L

(6)

To determine the kinetic energy of block A we need to have its speed. We know from kinematics that its speed is half that of the rope going over the pulley C. This rope ultimately attaches to mass B, which at State 2 is moving vertically downward. We can break vB at State 2 into two * * components, aligned along b 1 and b 2 :

We can see from geometry that |y| ˙ = |vB | sin β Thus we have |vB | sin β = 2|vA | SOLVE:

(7)

1 1 1 2 KE = mA ( vB sin β)2 + mB vB 2 2 2 2 m g PE = A (0.65L) 2 2

(7)→(3)⇒ (6)→(4)⇒ (1),(2),(8),(9)⇒

1 mB gL + kθ 2



2

π rad 2

1 1 2 2 sin2 β + mB vB + mA g(0.325)L = mA vB 8 2 2

1 π (10 kg)(9.81 m/s )(0.8 m) + (10 N·m/rad) rad 2 2 

2

2 vB



(5 kg) (10 kg) (sin(33.7))2 + 8 2



=

+ (5 kg)(9.81 m/s2 )(0.325)(0.8 m) 2 5.192 vB = 78.1 m2 /s2

vB = 3.88 m/s

140

(8) (9)

4.2.28 GOAL: Find safe length of bungie cord and impact velocity if cord is too weak. GIVEN: Initial and final height of jumper, mass of jumper and spring constant of bungie cord. DRAW:

FORMULATE EQUATIONS: Conservation of energy







KE + PE = KE + PE 1

1

2

(1)

2

SOLVE: (a) Kinetic and potential energies

PE g = mg(70 m),

1 1 KE = 0, PE g = mg(3 m), 2





KE = 0,

2

PE bc = 0

1 1 PE bc = 2 k(67 m − L)2 2

(2) (3)

Using (1) we get 1 (55 kg)(9.81 m/s2 )(70 m) = (55 kg)(9.81 m/s2 )(3 m) + (22 N/m)(67 m − L)2 2

(4)

L = 9.67 m (b) Kinetic and potential energies

KE = 0,

1 1 KE = 2 mv 2 , 2



PE g = mg(70 m),

1 PE g = 0, 2



PE bc = 0

1 1 PE bc = 2 k(70 m − L)2 2

(5) (6)

Again, using (1) we get 1 1 (55 kg)(9.81 m/s2 )(70 m) = (55 kg)v 2 + (0.9)(22 N/m)(70 m − 9.67 m)2 2 2 v = 7.95 m/s

141

(7)

4.2.29 GOAL: Find the maximum deflection of the spring after a moving mass strikes it. GIVEN: System parameter values. DRAW:

*

* ı  sin β − cos β cos β sin β

*

b1 b2

*

FORMULATE EQUATIONS: A force balance gives us *

*

*

N b 2 − S b 1 − mg *  = m¨ xb1 *

*

If the mass slides then S = µN . Resolving the equations of motion into the b 1 and b 2 directions yields: mg cos β − µN = m¨ x N − mg sin β = 0 Which, when combined, give us: m¨ x = mg(cos β − µ sin β) Thus, the force acting on the mass along the direction of the travel is mg(cos β − µ sin β). Work-energy gives us PE + KE + W1−2 = PE + KE 1

2

2

1

SOLVE: At top of the slide

KE = 0, 1



PE g = mgL cos β 1

At the instant of contact with the spring we have 1 KE = mv22 , 2 2



PE g = −mgL cos β, 2

W1−2 = −µN L

Invoking the work-energy between these two states 1 −µmgL sin β = mv22 − mgL cos β 2 √ 1 1 2 3 2 −0.1(9.81 m/s )(2 m) = v2 − (9.81 m/s )(2 m) 2 2 2 2

142

v2 = 5.66 m/s. At full compression we have

KE = 0, 3



PE g = 0, 3

1 PE s = k(∆x)2 2 3

Using work-energy between contact and full compression gives 1 1 mv22 − µmg sin β∆x = −mg cos β∆x + k(∆x)2 2 2 √ 1 3 1 2 1 2 2 (1.1 kg)v2 − 0.1(1.1 kg)(9.81 m/s ) ∆x = −(1.1 kg)(9.81 m/s ) ∆x + (3500 N/m)(∆x)2 2 2 2 2 (∆x)2 − (5.032 × 10−3 m)(∆x) − 1.006 × 10−2 m2 = 0 ∆x = 0.1029 m

143

4.2.30 GOAL: Determine θ˙ as a function of r and other constants. Determine r˙ as a function of r and other constants. Assuming L = 0, what is rmax for the trajectory? Verify part (c) by numerically integrating the equation of motion. GIVEN: System configuration and parameter values. DRAW:

(a): FORMULATE EQUATIONS: Apply conservation of angular momentum about O: H0 = mvD

(1)

˙ = mr2 θ˙ H0 = mr(rθ)

(2)

Angular momentum about O at A: Angular momentum at arbitrary position:

ASSUME: Because the spring force acts in a purely radial direction it won’t affect the angular momentum and thus we have (1), (2) ⇒

mr2 θ˙ = mvD

SOLVE: θ˙ =

vD r2

(3)

(b): FORMULATE EQUATIONS: Apply conservation of energy:    2 1 1 1 1 2 2 2 2 ˙ mv + k(D − L) = k(r − L) + m rθ + r˙ 2 2 2 2

(4)

SOLVE: D2 v2 mv + k(D − L) = k(r − L) + m + r˙ 2 r2 2

(3) → (4) ⇒ r˙ =

q

v 2 (r 2 −D2 ) r2

2

+

k m

!

2

(D2 − 2DL + 2rL − r2 )

(c): FORMULATE EQUATIONS: rmax occurs when preceding equation yields: 144

dr dt

= 0. Setting r˙ = 0 and L = 0 in the

 k  2 v2 r2 − D2 2 + D − r =0 r2 m


k v2 = 2 r m r2 = rmax = v

q

m k

v2m k

= 1.483 m

(d): FORMULATE EQUATIONS: The equations of motion are found from F = ma: m

h







i

r¨ − rθ˙2 e*r + rθ¨ + 2r˙ θ˙ e*θ = −k(r − L) e*r

e*r :

m(¨ r − rθ˙2 ) = −k(r − L)

(5)

e*θ :

rθ¨ + 2r˙ θ˙ = 0

(6)

ASSUME: There’s no need to integrate (6) because from conservation of angular momentum about O, we have vD θ˙ = 2 r

(7)

SOLVE: r¨ = −

(7)→(5)⇒

k (vD)2 (r − L) + m r3

(8)

Shown below is a segment of the data output from MATLAB after integrating the preceding equation. Note that at t = 0.2332 s we have r = 1.4832 m, a precise match to the theoretical prediction. The m-file code and a graphical plot of r(t) vs t is also shown.

t 0.1995 0.2062 0.2130 0.2197 0.2265 0.2332 0.2400 0.2467 0.2535 0.2602

r(t) 1.4498 1.4619 1.4713 1.4779 1.4819 1.4832 1.4817 1.4776 1.4707 1.4611

145

r(t) ˙ 1.9795 1.5874 1.1909 0.7910 0.3889 −0.0143 −0.4174 −0.8194 −1.2191 −1.6154

function dy=wk7p6(t,y) y1=y(1); y2=y(2); dy(1,1)=y(2); dy(2,1)=-50*y1/1.1+25/y1^3;

146

4.2.31 GOAL: a) Find the value of µ for which the mass will √ not stay in contact with the hoop GIVEN: m has an initial speed of vm = 3 gr = 9.905 m/s DRAW:

FORMULATE EQUATIONS: 



˙ e* = −mg * m (¨ r − rθ˙2 ) e*r + (rθ¨ + 2r˙ θ)  − N e*r − S e*θ θ

Force balance: 



˙ e* = e* (mg cos θ−N )+ e* (−mg sin θ−S) m (¨ r − rθ˙2 ) e*r + (rθ¨ + 2r˙ θ) θ r θ e*r :

m(¨ r − rθ˙2 ) = mg cos θ − N

e*θ :

˙ = −mg sin θ − S m(rθ¨ + 2r˙ θ)

GEOMETRIC CONSTRAINTS: r˙ = r¨ = 0 m(−rθ˙2 ) = mg cos θ − N mrθ¨ = −mg sin θ − S If the mass is moving in the positive θ direction, then S = µN . Hence we have: −mrθ˙2 = mg cos θ − N

(1)⇒ (1),(3)⇒

(1)

mrθ¨ = −mg sin θ − µN

(2)

N = mg cos θ + mrθ˙2

(3)

mrθ¨ = −mg sin θ − µ(mg cos θ + mrθ˙2 )

g (4) θ¨ + µθ˙2 + (µ cos θ + sin θ) = 0 r (4) is our equation of motion and (3) lets us determine the normal force between the mass and the hoop. (a) The minimum θ˙ value at θ = π rad can be found from (3) by setting N to zero: 147

2 0 = mg(−1) + mrθ˙min

θ˙min =

r

g r

To find the corresponding θ˙ value at θ = 0, we can apply conservation of energy: STATE 1: r g ˙ θ = π rad, θ = r STATE 2: θ=0







KE + PE = KE + PE 1

1

2

2

1 1 ˙ 2 2rmg + m(rg) = 0 + m(rθ) 2 2 5rg = r2 θ˙2 θ˙ =

q

5g at θ = 0 r

NUMERICAL: (b) For this part, we need to find the µ such that at θ = π, the normal force N goes to zero. θ˙ at θ = 0 is set to 9.905 rad/s. By using MATLAB and numerically integrating for a variety of µ values it as found that µ = 0.1555 causes the loss of contact to occur at θ = π.

148

4.2.32 GOAL: Find the maximum angle θ0 the boy can swing. GIVEN: System parameter values. DRAW:

*

*

er e*θ

* ı  − sin θ − cos θ − cos θ sin θ

ASSUME: The rope is inextensible and therefore its length is constant. Thus r = L and r˙ and r¨ are zero. FORMULATE EQUATIONS: A force balance gives us −T e*r − mg *  = m(−Lθ˙2 e*r + Lθ¨e*θ ) (mg cos θ − T ) e*r − mg sin θ e*θ = m(−Lθ˙2 e*r + Lθ¨e*θ ) The tension of the string does no work (it is perpendicular to the path). Therefore, conservation of energy holds KE + PE = KE + PE (1) 1

1

2

2

SOLVE: The maximum tension in the rope occurs at he bottom of the swing, where the potential energy is minimized and therefore the kinetic energy is at a maximum. Looking at the force balance in the e*r and e*θ directions for θ = 0 gives us θ¨ = 0 T = mLθ˙2 + mg

(2) (3)

Identifying State 1 as the system inclined at an angle θ and State 2 as the mass hanging straight down we have

KE = 0, 1



PE g = mgL(1 − cos θ)

1 1 ˙ KE = 2 m(Lθ2 )2 , 2

149

PE g = 0 2

(4) (5)

Substituting in (1) 1 mgL(1 − cos θ) = m(Lθ˙2 )2 2

(6)

We know that Tmax = (40 kg + 25 kg)(9.81 m/s2 ) = 637.7 N at θ = 0. Using (3) we get 637.7 N = (40 kg)(4 m)θ˙22 + (40 kg)(9.81 m/s2 ) Substitute in (6) and solving for the associated θ gives us θ1 = 46.6◦

150

⇒

θ˙22 = 1.53 s−2

(7)

4.2.33 GOAL: Find the maximum height, force on the center of mass, and speed at liftoff. GIVEN: You weigh 80 lb, you can jump 1 foot in a stationary elevator, your center of mass moves 6 inches, the elevator’s speed is 18 in/s. FORMULATE EQUATIONS: Stationary elevator: STATE 1: on the ground STATE 2: at the top of the leap PE + KE = PE + KE

0+

1 2

80 lb 32.2 ft/s2

2

1

1

!

v2 =









80 lb 32.2 ft/s2

2

!

(32.2 ft/s2 )(1 ft) + 0

v = 8 ft/s To attain a speed of 8 ft/s your legs muscles had to push against the force due to gravity. Assume your center of mass shifted 12 ft during the leap. Redefining our states as: STATE 1: Beginning of leg extension STATE 2: End of leg extension PE + KE + W1−2 = PE + KE



1

1 0+0+F = 2



1

80 lb 32.2 ft/s2

!



2

1 1 (32.2 ft/s ) + 2 2 2

 

2

80 lb 32.2 ft/s2

!

(8 ft/s)2

F = 239.0 lb Your legs produced 239 lb of force to launch you into your leap. Decelerating elevator: Now consider the leg extension and subsequent leap in a decelerating elevator. During the leg extension phase, we have: m(¨ y + z¨) = F − mg 80 lb 32.2 ft/s2

!

(¨ y − 12 ft/s2 ) = 239 lb − 80 lb y¨(t) = 76 ft/s2 y(t) ˙ = (76 ft/s2 )t y(t) = (76 ft/s2 )

y(t∗ ) =

1 (t∗ )2 = (76 ft/s2 ) 2 2 t∗ = 0.115 s

⇒ 151

t2 2

y(t ˙ ∗ ) = 76 ft/s2 (0.115 s) = 8.72 ft/s Your speed at the start of your leap is greater (8.72 ft/s) than in the case of a stationary elevator (8 ft/s). The difference between your speed and the decelerating elevator is given by: s˙ = (18 ft/s + 8.72 ft/s) − (32.2 ft/s2 )t − (18 ft/s − (12 ft/s)t) s˙ = 8.72 ft/s − (20.2 ft/s2 )t This equals zero (maximum separation) at : t=

8.72 ft/s = 0.432 s 20.2 ft/s2

152

4.2.34 GOAL: Find the normal force between the skater and the track. GIVEN: System configuration and parameters. DRAW:

FORMULATE EQUATIONS: State 1 corresponds to the skater stationary at a height of 16 ft above the ground and State 2 corresponds to the skater at a height for which θ = 45◦ . To determine the normal force, we need to know the skate boarder’s speed. We can find this from an energy approach. 1 h2 (t) = 10 ft(1 − √ ) = 2.929 ft 2 At State 1 all the energy is due to the gravitational potential: E = mgh1 = (70 lb)(16 ft) = 1120 lb·ft At State 2 we have both kinetic and gravitational energy: 1 35 lb E = mgh2 + mv 2 = (70 lb)(2.929 ft) + v2 2 32.2 ft/s2 Equating the energy at States 1 and 2 gives: 1120 lb·ft = 205 lb·ft +

35 lb v2 32.2 ft/s2

v = 29 ft/s Now that we have the velocity, we can determine the forces: m(rθ¨e* − rθ˙2 e* ) = −mg *  − N e* Force Balance: θ

e*r :

r

r

−mrθ˙2 = mg sin θ − N 70 lb 1 2 √ N= 2 (32.2 ft/s 2 32.2 ft/s 



+

(29 ft/s)2 ) 10 ft

N = 232.5 lb This is quite a bit of normal force. The skateboarder has to withstand over 3g’s at the halfway point down the curve.

153

4.2.35 GOAL: Analyze a bead/track interface. GIVEN: r = 0.4 m, each bead has a speed of 0.01 m/s. ASSUME: No friction between beads and wire. DRAW:

FORMULATE EQUATIONS: We know that at the midpoints, E and F , both beads must be moving at the same speed (from energy conservation) and in the same direction (both wires are oriented vertically). Thus the only variation can occur in the top half. For the right wire we have: y¨ = g y˙ = gt + v0 y=

gt2 + v0 t 2

We are given y(0) ˙ = 0.01 m/s and so have:

For the left wire we have: Force Balance:

y(t) ˙ = (9.81 m/s2 )t + 0.01 m/s

(1)

y(t) = (4.905 m/s2 )t2 + (0.01 m/s)t

(2)

−mg *  + N e*r = m(rθ¨e*θ − rθ˙2 e*r )

mrθ¨ = mg sin θ g θ¨ = sin θ r We can determine when the bead on the right reaches the midpoint from: gt2 r= + (0.01 m/s)t 2 e*θ :

154

(3)

t2 +

0.02 m/s 2(0.4 m) =0 2t − 9.81 m/s 9.81 m/s2

t2 + (2.04×10−3 s)t − 8.15×10−2 s2 = 0 t = 0.285 s We have to numerically integrate (3) to find the time for the bead to reach θ = π2 rad. The initial 0.01 m/s ˙ conditions are θ(0) = 0, θ(0) = 0.4 m = 0.025rad/s. Using ode45 in MATLAB yields a time of : t = 1.3098 s This is quite a bit longer than was the case for a purely vertical drop. Because the speed entry is the same for both beads at E and F , their behavior is identical from that point on. This analysis has shown that it takes a different amount of time for the bead to reach the same vertical position for the two cases. Now we’ll look at the bead’s speed. Because the kinetic energy for both cases will vary purely dependent upon the bead’s vertical position (the potential energy changes with vertical position and there is no friction to affect the energy balance) we would expect identical speeds when the bead’s are at the same vertical position. We can use the numerical integration results for the left wire and use the analytical results for the right one. A plot of v versus y for each case shows identical speed versus vertical position characteristics the two plots fall on top of each other. The velocities differ, due to the different orientation of the wire, but the speeds (and hence the kinetic energies) are the same.

155

4.2.36 GOAL: Show that KE2 = KE1 + W1−2 “works”. GIVEN: Form of the governing equation and the numerical approximation to use. FORMULATE EQUATIONS: (0.5 kg)¨ x = sin πt N

(1)

SOLVE: Shown at the end of this solution are the two M-files used in this problem. wk7p20.m just integrates (1). The output is a vector of times t and states y. They’re combined in the workspace into d: d[t,y]. After checking on the number of data points (n), n and d are used as inputs to wk7p20a.m. This computes the kinetic energies, which are plotted below. In the first plot, for which the work is given by F (tn )(x(tn+1 ) − x(tn )), there is somewhat reasonable agreement but a noticeable error near the end. This is because, even though the force is changing from x(tn ) to x(tn+1 ), the approximation treats it as fixed.

F (tn )+F (t

)

n+1 Letting the work be given by [x(tn+1 ) − x(tn )] corrects this error by using the average 2 F over each time interval. The corresponding plot is almost identical to the plot of 21 mx˙ 2 .

156

157

4.2.37 GOAL: Find the initial and final kinetic energy of the mass. GIVEN: Physical geometry and parameter values of the cockroach model. DRAW:

FORMULATE EQUATIONS:   Force balance, ˙ e* −mg *  − k(r − L) e*r − cr˙ e*r = m (¨ r − rθ˙2 ) e*r + (rθ¨ + 2r˙ θ) θ mass: −mg sin θ − k(r − L) − cr˙ = m(¨ r − rθ˙2 ) e* : r

˙ −mg cos θ = m(rθ¨ + 2r˙ θ)

e*θ :

(2) (3)

1 KE = mv 2 2

Kinetic Energy, mass:

(1)

(4)

SOLVE: 

d    dt 

(2)→ (3)⇒

r r˙ θ θ˙







r˙

   −g sin(θ) −   =    

k(r−L) m

−

θ˙ − gr cos(θ) −

cr˙ m

 + rθ˙2    

2r˙ θ˙ r

(5)

The initial conditions are r(0) = L

(6)

r(0) ˙ = v0 cos(θ0 ) π θ(0) = 3 − r˙0 tan(θ0 ) ˙ θ(0) = r0

(7) (8) (9)

Integrating using MATLAB (see code), the final speed is 0.1723 m/s. At θ =

π 3

, KE = 5 × 10−5 J, and at θ =

%Main File global g L c m k %Given g = 9.8; % m/s^2 L = .005; % m c = 1; % N*s/m m = 0.0025; % kg 158

2π 3

, KE = 3.71 × 10−5 J

k = 4000; % N/m v0 = -0.2; % m/s to the left tfinal = 1; %seconds %Calculate initial conditions r0 = L; % Spring initially unstretched theta0 = pi/3; % Initial touchdown angle rdot0 = v0/ (cos(theta0)+tan(theta0)*sin(theta0)); % Initial radial velocity m/s thetadot0 = -rdot0/r0*tan(theta0); % Initial angular velocity rad/s [t,y]=ode45(@BAM,[0 tfinal],[r0 rdot0 theta0 thetadot0]); r = y(:,1); rdot = y(:,2); theta = y(:,3); thetadot = y(:,4); stop = max(find(theta<2*pi/3)); %Find when angle = 2*pi/3; polar(theta(1:stop),r(1:stop)); vf = sqrt(rdot(stop)^2+ (r(stop)*thetadot(stop))^2); KEi = 1/2*m*v0^2 KEf = 1/2*m*vf^2 %------------------------------Separate File: BAM.m ------------------function xdot = BAM(t, x); global g L c m k %x(1) = r, x(2) = rdot, x(3) = theta, x(4) = thetadot % Differential Equation xdot(1) = x(2); xdot(3) = x(4); xdot(2) = -g*sin(x(3)) - k/m*(abs(x(1))-L)-c/m*x(2)+x(1)*x(4)^2; xdot(4) = -g/x(1)*cos(x(3)) - 2*x(2)*x(4)/x(1); xdot = [xdot(1); xdot(2); xdot(3); xdot(4)];

159

4.2.38 GOAL: Find the maximum height reached and the maximum compression of the pogo stick. GIVEN: Simplified pogo model and its parameter values. DRAW:

FORMULATE EQUATIONS:



State 1:

KE = 21 mv 2

PE = mgh0

State 2:

KE = 0

PE = mgh2

(2)

State 3:

KE = 0

1 PE = mgh3 + k|h3 − L|2 2 3

(3)

1

1



2

2



3

(1)

ASSUME: Total energy is conserved: E = KE + P E = Constant SOLVE:

(4)

1 mv 2 + mgh0 = mgh2 2

(1),(2),(4)⇒ h2 =

v2 (2.5 m/s)2 + 1.5 m + h0 = 2g 2(9.81 m/s2 )

h2 = 1.8 m (1),(3)⇒

1 1 mv 2 + mgh0 = mgh3 + k|h3 − L|2 2 2 h3 = 0.558 m

Numerically solving⇒ |h3 − L| = 0.642 m

160

(5) (6)

4.2.39 GOAL: Find angular velocity of the bowling ball at its lowest point. GIVEN: Geometry of arm/bowling ball model and parameters. DRAW:

FORMULATE EQUATIONS:







KE + PE + W1→2 = KE + PE 1

1

2

θf

Z

0 + mgL +

M dθ =

1 mv 2 + mgL(1 − sin θ) 2

0

GIVEN: θf = π/2 SOLVE: s 

(1)⇒

vf =

2 gL +

161

2

Mπ 2m



(1)

4.2.40 GOAL: Find the needed spring constant to allow Tarzan to reach Jane. Plot the Tarzan’s swing path up to reaching Jane. GIVEN: m1 = 81 kg, m2 = 60 kg, d = 20 m, h = 12 m, L = 15 m. DRAW:

ASSUME: Tarzan’s velocity vector should be oriented horizontally when reaching Jane’s position. GOVERNING EQUATIONS: Conservation of energy: PE + KE = PE + KE 1









1

2

2

SOLVE: Applying our energy balance gives us 1 1 1 mT g(12 m) + k(1 m)2 + 0 = mT v 2 + k(6 m)2 2 2 2 1 1 1 (81 kg)(9.81 m/s2 )(12 m) + k(1 m)2 + 0 = (81 kg)v 2 + k(6 m)2 2 2 2 9.535×103 N· m = (40.5 kg)v 2 + k(17.5 m2 ) Note that this is one equation in two unknowns. There exist an infinite number of solutions. Physically, they correspond to Tarzan at the level of the quicksand with a velocity vector that isn’t necessarily horizontal. If we confine ourselves to an energy approach, we won’t be able to determine the particular k for which Tarzan contacts Jane while moving horizontally. To solve the problem we need to consider the system kinetics.

162

*

*

er e*θ

* ı  − cos θ − sin θ sin θ − cos θ

The FBD=IRD diagram is shown along with a polar coordinate system. ˙ = −T e* − m g * mT (¨ r − rθ˙2 ) + mT (rθ¨ + 2r˙ θ)  Force balance: r T e*r :

mT (¨ r − rθ˙2 ) = −T + mT g sin θ

(1)

e*θ :

rθ¨ + 2r˙ θ˙ = g cos θ

(2)

(1) ⇒

r¨ = g sin θ −

k(r − 14 m) + rθ˙2 mT

g cos θ − 2r˙ θ˙ θ¨ = r We can put these equations into state form and integrate in MATLAB, using initial conditions of r = 15 m, r˙ = 0, θ = 0.562 rad and θ˙ = 0. The output data can be used to determine the x, y coordinates of Tarzan, using O (the attachment of the vine) as an origin: (2) ⇒

x = −r cos θ, y = 20 m − r sin θ By plotting the results, looking to see where the low point of the trajectory occurs and then altering k and iterating, we can ultimately find k = 292 N/m A plot of the trajectory is shown below.

163

(3)

(4)

4.2.41 GOAL: Determine the spring constant k needed to limit the compression of a restraining block to 5 cm. Determine as well how much a 1400 kg car would settle under a gravitation load if it had equivalent springs. GIVEN: mb = 9000 kg and v = 0.3 m/s. DRAW:

FORMULATE EQUATIONS: STATE 1: Just before collision, v = 0.3 m/s and the spring is uncompressed. STATE 2: At the maximum spring compression the boat has zero speed. The general energy conservation equation is





PE + KE = PE + KE



1

SOLVE:

1

2

2

1 1 (9000 kg) (0.3 m/s)2 = k(0.05 m)2 2 2 k = 3.24×105 N/m

If this spring was used on all four corners of a car we’d have a combined spring stiffness of 4k or 1.296×106 N/m Equating the force developed in the spring to counteract the force due to gravity gives us mg = kx (1400 kg)(9.81 m/s2 ) = (1.296×106 N/m)x x = 1.06×10−2 m

164

4.2.42 GOAL: Determine the maximum spring compression after a mass/spring combination strikes a wall. GIVEN: v = 40 mph, weight = 3000 lb, k = 80, 000 lb/ft DRAW:

FORMULATE EQUATIONS: STATE 1: Just before collision, x˙ = 40 mph, spring uncompressed. STATE 2: At maximum spring compression (∆x), x˙ = 0.







PE + KE = PE + KE 1

1

2

2

SOLVE: 1 2

3000 lb 32.2 ft/s2

!

88 ft/s 40 mph 60 mph 

2

∆x = 2.0 ft

165

1 = (80, 000 lb/ft)∆x2 2

4.2.43 GOAL: Determine if a falling spring restrained person will hit the ground. GIVEN: Spring constant, mass, initial height and dimensions DRAW:

FORMULATE EQUATIONS: We’ll apply conservation of energy: 1

g 1

+ PE

sp

= KE + PE 2

1













KE + PE

g 2

+ PE

sp

2

SOLVE: Assume that initially her speed is zero.

KE = 0 1

PE

g 1

= mgh1 = (57 kg)(9.81 m/s2 )(50 m) = 2.80 × 104 N·m

PE

sp

=0 1

Solve for her speed after falling 50 m 1 KE = mv22 = (28.5 kg)v22 2 2

PE

PE

sp

2

g 2

=0

1 1 = kx2 = (63 N/m)(50 m − 20 m)2 = 2.84 × 104 N·m 2 2 2.80 × 104 N·m = (28.5 kg)v22 + 2.84 × 104 N·m v22 = −14(m/s)2

A negative value for v22 implies an imaginary solution for v2 . The conclusion is that she doesn’t contact the ground, but rather comes to rest somewhat above the ground.

166

4.2.44 GOAL: Find Favg during impact and formulate an expression for the minimum spring length GIVEN: Physical dimensions of the system DRAW:

FORMULATE EQUATIONS: Non-conservative work is done during the drop and thus we’ll need to account for this in our energy balance: KE1 + P Eg1 + P ESP 1 − F ∆x = KE2 + P Eg2 + P ESP 2

(1)

SOLVE: a) Go from state 1 to just before impact (state 2, B moving at vB ) m g 1 mB vo2 + mB gh − 2Ff h − d − L − A 2 k 

2 = vB





m g 1 2 = mB vB + mB g L − A + d 2 k 

m g 2 1 mB vo2 + mB gh − 2Ff h − d − L − A mB 2 k 









− mB g L −

mA g +d k





(2)

After impact we have 



mB vB + mA vA = mA + mB v Using vA = 0 gives v=

mB mA + mB

!

vB

(3)

Impulse implies: mB vB − F ∆t = mB v ⇒ F =

m v −m v B B B ∆t

Where vB and v are found from (2) and (3) b) Now we’ll go from just after impact, with A and B moving with speed v, to vA = vB = 0  m g m g mA g 1 1 mA + mB v 2 +mA g L − A +mB g L − A + d + k 2 k k 2 k 







167



2

m g L − A − Lmin k



−2Ff



  1  2 = mA gLmin + mB g Lmin + d + k L − Lmin 2 This can then be solved for Lmin

168

4.2.45 GOAL: Find the average force acting between two mass particles during a collision and expressions for the distance traveled by the post-collision pair under different frictional conditions. GIVEN: System configuration. DRAW:

FORMULATE EQUATIONS: We’ll apply conservation of energy







KE + PE = KE + PE 2

1

1

2

to determine the collision speed vc , conservation of momentum to calculate the post-collision speed vpc and work/energy







KE + PE + W1−2 = KE + PE 1

1

2

2

to calculate the distance traveled by the mass-pair. SOLVE: The spring is stretched to the same degree when m1 is released and when it contacts m2 . Thus it needn’t be included in our energy balance. Going from release to just before collision we have 1 1 m v 2 + m1 gr = m1 vc2 2 1 0 2 vc =

q

v02 + 2gr

Conservation of linear momentum then yields m1 vc = (m1 + m2 )vpc vpc =

m1 m1 + m2

!

vc

*

We know that the applied linear impulse (F∆t) acts to speed up m2 and slow down m1 . We can examine either mass to find |F |. Let’s choose m2 . In the * ı direction we have   m m vc F ∆t = m2 vpc − 0 = 1 2 m1 + m2 Hence we have |F | =

m m vc 1 2
m +m ∆t 1 2

Next we need to examine how far the two masses slide. Because the resisting force is a constant, the work done is simply the force multiplied by the displacement. Note that we’ll define x = 0 to be the initial position of m2 and x is positive to the right ( * ı direction). 1 2 k(r

2 − F ∆x = 1 k − L)2 + 21 (m1 + m2 )vpc c 2

169

hp

i2

r2 + (∆x)2 − L

This is a single equation with the single unknown ∆x. Now consider a more complicated resisting force, that of Coulomb friction (F = µd N ).

The friction force µd N varies with N and N varies with the position of the two masses: N − (m1 + m2 )g + T sin θ = 0 ⇒ N = (m1 + m2 )g − T sin θ The spring force T is given by T =k and sin θ is given by

√ r . r 2 +x2

q



r2 + (∆x)2 − L

Hence we have N = (m1 + m2 )g − kr + √

kLr x2 + r2

The force due to friction is µd N and so our energy equation becomes 1 2 k(r

2 − − L)2 + 21 (m1 + m2 )vpc

R ∆x 0

µd N dx = 21 k

hp

i2

r2 + (∆x)2 − L

This is again one equation in one unknown and can therefore be solved numerically for ∆x.

170

4.2.46 GOAL: Find the impact speed of the pile driver with the ground. GIVEN: Mass of pile driver and dimensions, mass and orientation of support structure. DRAW:

FORMULATE EQUATIONS: We can use conservation of energy: STATE 1: KE1 = 0 P E1 = mgL1 cos 30◦ STATE 2:

1 KE = mv 2 2 2



PE = 0 2

SOLVE:

1 mgL1 cos 30◦ = mv 2 2 √

s

v=

q

2gL1 cos 30◦ =

171

2

2(9.81 m/s )(2 m)(

3 ) = 5.83 m/s 2

4.2.47 GOAL: Find the speed of block A when it loses contact with the horizontal surface its sliding upon. GIVEN: System configuration, masses and spring/force characteristic. DRAW:

*

*

b1 b2

*

* ı  cos θ sin θ − sin θ cos θ

FORMULATE EQUATIONS: We’ll apply energy conservation. First, we need to determine how toaccount for the spring’s potential energy. Comparing the force characteristic F = 5mg L − L0 with the characteristic of a linear spring (F = kx) lets us identify the spring’s L 0

potential energy as

PE

sp

=

2 1 5mg  L − L0 2 L0

where the linear spring’s potential energy is given by 21 kx2 . The two blocks are connected by an inextensible spring and therefore we know that they both have the same speed. Assuming that block B falls a distance ∆x when contact is lost, we start with a potential energy of mg∆x, zero kinetic and zero potential energy. When block B has fallen ∆x, both blocks are moving with speed v and the spring is stretched to a length L from its initial length L0 : 1 1 5mg 1 mg∆x = mv 2 + mv 2 + (L − L0 )2 (1) 2 2 2 L0 Our force condition for a loss of contact is given by 5mg (L − L0 ) cos θ = mg L0 SOLVE: L=

(2)⇒ (1) ⇒ (3)⇒

mv 2 +

(2)

5L0 4

(3)

1 5mg (L − L0 )2 = mg∆x 2 L0

∆x = 172

q

L2 − L20 =

3L0 4

(4) (5)

2

(5)→(4)⇒ (6)⇒

mv 2 +

3L 1 5mg L0 = mg 0 2 L0 16 4

v=

173

q

19 32 gL0

(6)

4.3

Power and Efficiency

174

4.3.1 GOAL: Find the power output of a person climbing stairs. GIVEN: Rate of climb, height of a stair, number stairs per flight and flights per floor. DRAW:

FORMULATE EQUATIONS: For a constant power applied over a time t we have the relationship P = W/t where W is the work done. SOLVE: The force being applied by the climber is simply N = mg because the climbing is being done at a constant speed. The change in height h over a single floor is given by h = (7 in/step)(13 steps/flight)(2 flights/floor) = 182 in = 15.16 ft P =

(168 lb)(15.16 ft) W = t 12 s P = 212 lb· ft/s = 0.39 hp

175

4.3.2 GOAL: Find the cost of raising an elevator and one passenger up 4 stories and the average power required. GIVEN: Elevator efficiency, cost of electricity, mass of elevator/passenger, distance traveled and time needed for the trip. DRAW:

FORMULATE EQUATIONS: For a constant power applied over a time t we have the relationship P = W/t where W is the work done and for an efficiency η we have Pout = ηPin , Wout = ηWin SOLVE: The work done is equal to the force times the distance moved: W = (270 kg + 70 kg)(9.81 m/s2 )(19 m) = 6.34×104 N· m = 6.34×10 kJ Given an efficiency of 0.84 this means that the input work must have been 6.34×10 kJ Win = = 7.54×10 kJ 0.84 The average power used is therefore W 7.54×10 kJ P = = = 3.77 kW t 20 s The cost of electricity used is given by cost = ($0.12/ kW· hr)(1 hr/3600 s)(20 s)(3.77 kW) = $0.0025 cost = 0.25 cents

176

4.3.3 GOAL: Find cost to fill a water tank. GIVEN: Height of tank, quantity of water, efficiency of pump. DRAW:

FORMULATE EQUATIONS: W1−2 =

mgh (1000 kg)(9.81 m/s2 )(30 m) = = 5.89 × 105 N·m (0.5) (0.5)

SOLVE: 1 kW · hr = (1000 N · m/s)(3600 s) = 3.6 × 106 N·m # kW · hr needed =

5.89 × 105 N·m = 0.16 3.6 × 106 N·m

Cost = (0.16)($0.15) = 2.5 cents

177

4.3.4 GOAL: Find the power dissipated due to braking. GIVEN: Slope, weight of car, speed and acceleration. DRAW:

*

*

b1 b2

*

* ı  cos θ sin θ − sin θ cos θ

θ = tan−1 0.1 = 5.71◦ FORMULATE EQUATIONS: 30 mph = 44 ft/s We’ll use our power/speed relationship to determine the power: P = F s˙ *

Force balance:

*

*

m¨ s b 1 = −mg *  + F b1 + N b2

*

m¨ s = −mg sin θ + F

b1 :

*

(1)

0 = −mg cos θ + N (2) b2 : SOLVE: The power dissipated by the brakes is equal to the force exerted on the car, F , times the car’s speed, s. ˙ With s¨ equal to zero we have 0 = −mg sin θ + F ⇒ F = mg sin θ P = F v = (3800 lb) [sin (5.71◦ )] (44 ft/s) = 1.66 x 104 lb·ft/s = 30 hp

178

4.3.5 GOAL: Find the power dissipated due to braking GIVEN: Slope, weight of car, speed and acceleration DRAW:

*

*

b1 b2

*

* ı  cos θ − sin θ sin θ cos θ

ASSUME: Motion remains along the hill’s surface FORMULATE EQUATIONS: P = Fv *

Force balance:

*

*

m¨ s b 1 = −mg *  − F b1 + N b2

*

b1 :

m¨ s = mg sin θ − F

(1)

*

0 = −mg cos θ + N

(2)

b2 :

θ = tan−1 (0.08) = 4.57◦ SOLVE: The power dissipated by the brakes is equal to the force exerted on the car, F , times the car’s speed, s. ˙ Initially s¨ = 0 and so F = mg sin θ. Call this F1 . When the car begins to decelerate at 0.8g we have m (−0.8g) = mg sin θ − F ⇒ F = mg (sin θ + 0.8) Denote this value of F as F2 . Initial Power = F1 v = (3200 lb) (sin (4.57◦ )) (44 ft/s) = 1.12 x 104 lb·ft/s Final Power = F2 v = (3200 lb) (sin (4.57◦ ) + 0.8) (44 ft/s) = 1.24 x 105 lb·ft/s Change in power is 1.24 x 105 − 1.12 x 104 ⇒ 1000% change 1.12 x 104

179

4.3.6 GOAL: Find how far up a mountain a single container of yoghurt will get you. GIVEN: Body’s effiiciency, mass of the body/bicycle and calories in yoghurt. DRAW:

FORMULATE EQUATIONS: For an efficiency η we have Wout = ηWin SOLVE: The work done is equal to the force times the distance moved. We neglect all external drag sources and simply consider the work done to elevate the body. Note that 1 “food” calorie is actually equal to 1000 calories (the calories for which 1 calorie=4.184 J). We’ll start with the energy supplied by the yoghurt: 4, 184 J E = 150 food cal = (150 food cal) = 627.6 kJ 1 food cal We’re given that the body is 25% efficient and thus we will only extract (0.25)(627.6 kJ) = 156.9 kJ from it. Equating the work done with the energy available yields F h = (75 kg)(9.81 m/s2 )h = 156.9 kJ h = 213 m

180

4.3.7 GOAL: Find the power output needed to go up a specified height in a specified time. GIVEN: Cyclist’s weight, height attained and travel time. DRAW:

FORMULATE EQUATIONS: We’ll be using our power/work relationship for constant work done: W P = t SOLVE: (160 lb)(3500 ft) Fh = = 212.1 ft· lb/s P = t (44 min)(60 s/1 min) P = (212.1 ft· lb/s)

181

(1 hp) = 0.386 hp (550 ft· lb/s)

4.3.8 GOAL: Find the average power put out in climbing a given hill at a certain rate and then determine the change in time if an electric motor is added. GIVEN: Mass of the system, height of the mountain and initial time needed to reach the top. DRAW:

FORMULATE EQUATIONS: We’ll be using our power/work relationship for constant work done: W P = t SOLVE: Fh (175 lb)(3500 ft) P = = = 136.1 ft· lb/s t (75 min)(60 s/1 min) P = (136.1 ft· lb/s)

(1 hp) = 0.247 hp (550 ft· lb/s)

Now we can modify our system by adding a 20 lb, 1/3 hp motor. This increases the overall hp to 0.581 and the weight to 195 lb. W (195 lb)(3500 ft) t= = 550 ft· lb/s P (0.581 hp) 1 hp t = 2137 s = 35.6 min The addition of the electric motor would get me to the top 39.4 minutes faster.

182

4.3.9 GOAL: Determine the coefficient a (part of a drag force formula). Also determine the change in speed associated with a 10% increase in power. GIVEN: Form of the drag force, weight and speed of the cyclist and steady power output. DRAW:

FORMULATE EQUATIONS: Our diagram only shows the forces along the direction of travel, the traction force T and the drag force Fd . We’ll be using our power/speed relationship: P = Fv SOLVE: In a steady-state condition we have a balance between the drag force and the traction force: T = Ff = av 2 Using this traction force in our power expression gives us P = av 3 We’re given that the power output is 0.35 hp and therefore can solve for a: 0.35 hp = 192.5 ft· lb/s = a(36.6 ft/s)3 a=

192.5 ft· lb/s (36.6 ft/s)3

a = 3.90×10−3 lb· s2 / ft2 Increasing the power output by 10%, to 0.385 hp means we have 0.385 hp = 211.75 ft· lb/s = (3.90×10−3 lb· s2 / ft2 )v 3 ⇒ v = 37.9 ft/s = 25.8 mph The change in the cyclist’s speed is just 0.8 mph , a bit over a 3 percent increase in speed for a 10 percent increase in power.

183

4.3.10 GOAL: Find the percentage increase in power output for double the speed. GIVEN: Relation between aerodynamic drag and speed. DRAW

ASSUME: Let’s assume that, for the two cases in which we’re interested, the force exerted by the body exactly matches the drag force, such that it is not accelerating. FORMULATE EQUATIONS: Drag:

Fd = av 2

(1)

Power:

P = Fd v

(2)

SOLVE: Let v1 be the initial velocity, and v2 be double the initial velocity: v2 = 2v1 . The expressions for power at these two states are P1 = Fd v1 = av13 P2 = Fd v2 = a(2v1 )3 The percentage increase is given by 7av13 a(8v13 − v13 ) P2 − P1 × 100 = × 100 = 700% × 100 = % increase = P1 av13 av13 % increase = 700%

184

4.3.11 GOAL: Find the power output needed to drive up a grade at a constant speed. GIVEN: m = 1200 kg, 5% grade, v = 20 m/s DRAW:

*

*

b1 b2

*

* ı  cos θ sin θ − sin θ cos θ

θ = tan−1 (0.05) = 2.86◦ ASSUME: Neglect air and road drag FORMULATE EQUATIONS: *

P = F · v*

Power FBD=IRD SOLVE: * b1 :

*

*

N b 2 + F b 1 − mg *  = m¨ x=0 F − mg sin θ = 0 ⇒ F = mg sin θ F = mg sin θ = mgθ = mg(0.0499) *

*

*

P = F · v* = mg(0.0499) b 1 · v b 1 P = (1200 kg)(9.81 m/s2 )(0.0499)(20 m/s) = 11.7 kW P = 11.7 kW = 15.8 hp

185

4.3.12 GOAL: Find new velocity when grade increases. GIVEN: Constant power condition and grade change. Initial speed is 10 mph. DRAW:

FORMULATE EQUATIONS: We’ll apply work energy:







KE + PE = KE + PE + Wnc1−2 2

(1)

1

1

2

P =

Definition of power:

dW dt

(2)

The two angles of ascent are given by θ1 = arctan 0.05 = 2.86◦ , θ2 = arctan 0.06 = 3.43◦ Let the speed on the initial slope be given by v1 and on the greater slope be given by v2 . SOLVE: 0 + mgh = 0 + 0 + Wcyclist (1) ⇒ (2), (3) ⇒

P = 

dh dt

dWcyclist dt



 1

= v1 sin θ1 ,

= mg

dh dt

(4)

dh dt

= v2 sin θ2

(5)



P1 = P2 ⇒ v2 = v1

(2), (4), (5) ⇒

2

sin θ1 sin θ2

v2 = 8.34 mph

186

(3)

(6)

4.3.13 GOAL: Find the maximum attainable speed of the car and the horsepower at that speed. GIVEN: The car weighs 3500 lb, cr = 0.02, cr = 0.011 slug/ft, and F = 350 lb FORMULATE EQUATIONS: At the maximum speed x ¨ = 0. Thus a force balance gives us 0 = −cr mg − k(x) ˙ 2+F 0 = −(0.02)(3500 lb) − (0.11 slug/ft)(x) ˙ 2 + 350 lb x˙ = 160 ft/s = 109 mph At this speed, the power is given by: P = F v = (350 lb)(160 ft/s) = 56, 000 lb·ft/s = 102 hp

187

4.3.14 GOAL: Find the average horsepower generated by the cars. GIVEN: The cars are initially at rest and accelerate to 15 mph. µd = 0.55, µs = 0.8, and both cars weigh 1500 lb. ASSUME: 35% of the weight is supported by the rear wheels and 65% is supported by the front wheels. FORMULATE EQUATIONS: 3000lb = 93.2 slg, 15 mph = 22 ft/s mA = mB = 32.2 ft/s2 Car A:

Fd

= 0.8(3000 lb)(0.65) = 1560 lb (drive force)

Car B:

Fd

= 0.55(3000 lb)(0.65) = 1073 lb (drive force)

A

B

Apply force balance Car A: (93.2 slg)¨ xA = 1560 lb x ¨A = 16.7 ft/s2 To reach 15 mph (22 ft/s) requires: (16.7 ft/s2 )(tA ) = 22 ft/s tA = 1.31 s hpA = F v = (1560 lb)(22 ft/s) = 34, 320 lb/ft = 62.4hp hp hpavg = 62.4 1.31 s = 47.5 hp Apply a force balance to Car B: (93.2 slug)¨ xB = 1073 lb x ¨B = 11.5 ft/s2 To reach 15 mph (22 ft/s) requires: (11.5 ft/s2 )(tB ) = 22 ft/s tB = 1.91 s hpB = F v = (1073 lb)(11.5 ft/s) = 12340 lb/ft = 22.4 hp hp hpavg = 22.4 1.91 s = 12 hp Car B takes more than 50% as long as car A to reach 15 mph. It seems that slipping is a far less effective way of transferring power from the engine to the car than no-slip. The average horsepower for the slip case is lower both because the force is lower (1073 lb vs 1560 lb) and the time over which the average was taken is longer. 188

4.3.15 GOAL: What is the horsepower developed after 3 seconds? What is the average horsepower over the distance traveled? DRAW:

FORMULATE EQUATIONS: Force balance, * ı:

m¨ x=F

(1)

P = F x˙ (2) Power: SOLVE: If the vehicle is rolling without slipping then the force sustainable by the frictional interface is Fdrive = mgµs = (1500 kg)(9.81 m/s2 )(0.9) = 13, 244 N This implies an acceleration of (1500 kg)¨ x = Fdrive = 13, 244 N x ¨ = 8.83 m/s2 After 3 seconds the vehicle is traveling at x˙ = (8.83 m/s2 )(3 s) = 26.5 m/s The power is equal to the force times the velocity: P = F x˙ = (13244 N)(26.5 m/s) = 351 kW In terms of horsepower this is 1 hp P = (351 × 10 W) 746 W 3





= 470 hp

P = 470 hp x˙ increases linearly with time. The average hp is therefore hpavg = 470 2 = 235 hp

189

4.3.16 GOAL: Determine a motor’s effiiciency. GIVEN: Mass of load, speed of load, slope’s inclination and electrical power input. DRAW:

FORMULATE EQUATIONS: We’ll use our FBD to determine the tension applied to the load and then use P = T v to the find the power and η = Pout /Pin to find the efficiency. SOLVE: From the FBD we have T − mg sin θ = 0 T = mg sin θ = (1000 kg)(9.81 m/s2 )(0.5) = 4905 N P = T v = (4905 N)(2 m/s) = 9810 N· m/s = 9810 W η=

9810 W = 0.545 18×103 W

Thus we see that η = 0.545 and we have an efficiency of 54.5 percent .

190

4.3.17 GOAL: Find the amount of power and work required to move gravel. GIVEN: Hopper moves at 1.5 m/s. DRAW:

FORMULATE EQUATIONS: We’ll use the power/speed formulation P = Fv SOLVE: θ = tan−1

8 ft = 16.5◦ 22 ft

P = F v = (100 lb + 30 lb)(1.5 ft/s) cos(16.5◦ ) = 187 lb·ft/s P = 187 lb·ft/s = 0.34 hp W = F ∆x = (130 lb)(27 ft) = 3510 lb·ft

191

4.3.18 GOAL: Find the input power to ride at steady speed and the effect of road grade on speed. GIVEN: Drag force and cyclist’s efficiency (a) DRAW:

FORMULATE EQUATIONS: Newton’s law decomposed in the * ı and *  directions FT − a1 − a2 v 2 = m¨ x N − mg = 0

(1) (2)

SOLVE: For a constant speed x ¨ = 0. Substituting in (1) gives us FT = a1 + a2 v 2 The normal force N does no work so we need only consider FT . Velocity is given by v = 25 mph = 11.2 m/s. The power output is Pout = FT v = (3.3 N + (0.24 N· s2 /m2 )(11.2)2 )(11.2 m/s) = 372 W

(3)

The efficiency is 96% so the power input is Pin =

372 W = 387 W = 0.52 hp 0.96

(4)

(b) DRAW:

*

*

FORMULATE EQUATIONS: A force balance decomposed in the b 1 and b 2 directions gives us FT − mg sin θ − a1 − a2 v 2 = 0 N − mg = 0

(5) (6)

We’re looking for a steady speed and thus the acceleration is zero. SOLVE: Use (5) to find FT = mg sin θ + a1 + a2 v 2 = (85 kg)(9.81 m/s2 )(0.05) + 3.3 N + (0.24 N· s2 /m2 )v 2 = 44.9 N + 0.24v 2 (7) 192

From part (a) we know the power driving the bike/person up the hill is 372 W. 372 W = (44.9 N + (0.24 N· s2 /m2 )v 2 )v

(8)

v = 6.68 m/s = 14.9 mph This is a huge decrease in speed and highlights the well known fact that when you ride up a hill you go a lot slower that when you’re riding on the flats.

193

4.3.19 GOAL: Find average power output. GIVEN: Average speed is 11 mph, the road has an average grade of 5.5%, m = 76 kg, road drag force is 2.5 N DRAW:

FORMULATE EQUATIONS: *

Force balance, b 1 :

Ftraction − Fdrag − mg sin θ = 0 sin θ = sin(tan−1 (0.055)) = 0.055 rad 11 mph = 17.7 kph = 4.92 m/s Ftraction = mg sin θ + 2.5 N + (0.2 N·s2 /m2 )v 2 Ftraction = (9.81 m/s)(76 kg)(0.055 rad)+2.5 N+(0.2 N·s2 /m2 )(4.92 m/s)2

P ower = Ftraction v = (48.3 N)(4.92 m/s) = 237 W = 0.32 hp

194

4.3.20 GOAL: DRAW:

FORMULATE EQUATIONS: F − Fdrag − mg sin θ = 0

*

Force balance, b 1 :

F = mg sin θ + 2.5 N + (0.21 N· s2 /m2 )v 2 F = (9.81 m/s2 )(76 kg)(0.06) + 2.5 N + (0.21 N· s2 /m2 )v 2 F = 47.2 N + (0.21 N· s2 /m2 )v 2 P = Fv = (47.2 N + (0.21 N· s2 /m2 )v 2 )v We can convert our power output from hp to watts: 0.31 hp = 0.31 hp(746 W/hp) = 231 W 231 W = (47.2 N)v + (0.21 N· s2 /m2 )v 3 (0.21 N· s2 /m2 )v 3 + (47.2 N)v − 231 W = 0 v = 4.50 m/s Now consider the case of no road drag at all. F = mg sin θ + 0.21v 2 = (0.21 N· s2 /m2 )v 2 + 44.7 N 231 W = [44.7 N + (0.21 N· s2 /m2 )v 2 ]v (0.21 N· s2 /m2 )v 3 + (44.7 N)v − 231 W = 0 v = 4.69 m/s Eliminating all road drag increases my velocity by 4.3% 4.3% increase

195

4.3.21 GOAL: Determine an electric motor’s efficiency. GIVEN: Mass of load, distance traveled and time needed to lift load and electrical power supplied to the motor. DRAW:

FORMULATE EQUATIONS: We’ll determine the work done (force applied over distance), then use P = W/t to find the average power and finally use η = Pout /Pin to determine the efficiency. SOLVE: The work done is given by W = T h = (200 kg)(9.81 m/s2 )(4 m) = 7848 J W 7848 J = = 2616 W t 3s 2616 W η= = 0.872 3000 W

P =

Thus we see that η = 0.872 and we have an efficiency of 87.2 percent .

196

4.3.22 GOAL: Plot the power acting on a block. GIVEN: Mass of block, velocity profile and inclination of slope. DRAW:

FORMULATE EQUATIONS: We’re asked to find the overall power applied to the block. Two forces are acting - gravity and the tension in the rope. Because we’re given the acceleration of the block, we don’t actually need to distinguish between the two. We know from a force balance that F = m¨ s where F represents the sum of the forces acting in the direction of motion. Using this force in P = F v will allow us to determine the power. SOLVE:   2 2 F = m¨ s = (125 kg) m/s = 83.3 N Force balance: 3 The power acting on the block is given by P = (83.3 N)s˙ We can see that the power is a linearly increasing function of s. ˙ The acceleration of the block is constant and thus we have   2 m/s2 t s˙ = s¨t = 3 and

2 m/s2 t = (55.5 W/s)t P = (83.3 N) 3 

197



198

4.3.23 GOAL: Determine the zero to 30 mph time for a car that has no drag forces and puts out its maximum power at all times. GIVEN: Mass of the car and the maximum horsepower of its engine. DRAW:

FORMULATE EQUATIONS: To determine the time needed we’ll use our speed/time/constant power relationship mv 2 ∆t = 2P where v is the final speed (car starts from zero) and P is the applied power. SOLVE: 3200 lb m= = 99.4 slg 32.2 ft/s2 P = 200 hp = (200 hp) ∆t =

550 ft· lb/s = 110, 000 ft· lb/s 1 hp

mv 2 (99.4 slg)(44 ft/s2 )2 = = 0.875 s 2P 2(110, 000 ft· lb/s ∆t = 0.875 s

199

4.3.24 GOAL: Plot the rate at which energy is removed from a body that’s slowing at a specified rate. GIVEN: Mass of the body and time to slow to zero. DRAW:

FORMULATE EQUATIONS: We’ll use the power/time relationship P = Fb v SOLVE: The body slows from 100 km/hr to zero in 4 s, which implies a constant acceleration of 0 − 27.7 m/s a= = −6.94 m/s2 4s Fb = ma = (1600 kg)(−6.94 m/s2 ) = −11, 104 N P = Fb v = (−11, 104 N)v

200

4.3.25 GOAL: Determine the drag coefficient cd acting on a car. GIVEN: Relevant efficiencies, fuel usage and energy density of gasoline. DRAW:

FORMULATE EQUATIONS: What we’ll do is compute the total power in a gallon of gasoline. We’ll then reduce this by the combined drivetrain and combustion efficiency (ηdt and ηcom , respectively) to get the total work supplied to move the car. We’ll next take our expression for the force acting on the car (F ) that counteracts the drag force. Multiplying by the distance traveled will give us the work. Equating these two energies will allow us to compute cd . SOLVE: The total energy in a gallon of gasoline is 1.25×108 J and the delivered energy is Edel = (1.25×108 J)ηdt ηcom = (1.25×108 J)(0.8)(0.28) = 2.8×107 J The car burns 2.7 gal/hr and thus delivers a total energy of E = 2.7(2.8×107 J) = 7.56×107 J The energy used by the car is given by the force times the distance traveled. In one hour we’ll use 2.7 gal of gasoline and travel 60 miles = 9.66×104 m. The work done is thus W = F d = 0.5ρAcd v 2 d = 0.5(1.2 kg/m3 )(2 m2 )cd (26.8 m/s)2 (9.66×104 m) W = cd (8.34×107 kg· m2 /s2 ) Equating the work done and the energy used gives us 7.56×107 J = cd (8.34×107 kg· m2 /s2 ) cd = 0.91

201

4.3.26 GOAL: Determine power supplied to an electric motor. GIVEN: Mass of load, distance traveled and time needed to lift load and motor’s efficiency. DRAW:

FORMULATE EQUATIONS: We’ll determine the work done (force applied over distance), then use P = W/t to find the average power and finally use η = Pout /Pin to determine the input power given the known efficiency. SOLVE: Two ropes support the mass, giving us an equilibrium (zero acceleration) condition of mg 2T = mg ⇒ T = 2 We know from our pulley kinematics that the reel will pull rope into it at twice the speed with which the load is lifted. Thus, if the load moves up 3 m in 5 s, its speed is 0.6 m/s and the rope feeds into the reel at 1.2 m/s. The power done at the reel is given by (50 kg)(9.81 m/s2 ) P = Tv = (1.2 m/s) = 2.94×102 W 2 η = 0.60 =

2.94×102 W ⇒ Pin = 4.91×102 W Pin Pin = 4.91×102 W

202

4.3.27 GOAL: Determine the maximum acceleration you can achieve on your bicycle. Determine the time needed to catch up to other person and distance traveled. GIVEN: Mass of you and the bicycle, the drag forces acting and your maximum power output. FORMULATE EQUATIONS: (F − Fdrag ) * ı + (N − mg) *  = m¨ x* ı

Force balance:

F − Fdrag = m¨ x

*

ı:

⇒

x ¨=

1 (F − Fdrag ) m

(1)

(a): SOLVE: In steady state conditions we have: F = Fdrag = 2.8 N + (0.22 N·s2 / m2 )v 2

(2),(3)⇒

(2)

v = 15 mph = 24.2 kph = 6.7 m/s

(3)

F = 2.8 N + (0.22 N·s2 /m2 )(6.7 m/s)2 = 12.7 N

(4)

P = F v = (12.7)(6.7) = 85 W Steady state power is 85 W Maximum power is 0.4 hp = (0.4 hp)(745.7 W/hp) = 298 W. At 6.7 m/s this implies a force of F (6.7 m/s) = 298 W 

(5)→(1)⇒

x ¨=

1 82 kg



⇒

44.5 N

(5)

(44.5 N − 12.7 N) = 0.388 m/s2

(b): To determine the positions and speed versus time we need the governing equation for the bicycle. The maximum force the cyclist can produce is given by Fmax x˙ = 298 W ⇒ F =

298 W x˙

Our equation of motion is therefore given by 1 1 x ¨= (F − Fdrag ) = 82 kg max 82 kg



298 W − 2.8 N − (0.22 N·s2 /m2 )x˙ 2 x˙



The following shows a simple MATLAB file that contains this equation as well as plots of x and x˙ versus time. The time needed to reach 20 mph is 8.86 s.

203

204

4.3.28 GOAL: Plot the rate at which energy is removed from a body that’s slowing at a specified rate. GIVEN: Mass of the body and distance needed for the deceleration. DRAW:

FORMULATE EQUATIONS: We’ll use the power/speed relationship P = Fb v and our formula that relates distance traveled, speed and acceleration (under a constant acceleration assumption):  1 2 v2 − v12 (1) a(x2 − x1 ) = 2 SOLVE: The body slows from 60 mph (88 ft/s) to zero in 120 ft with a!constant deceleration. 3500 lb m= = 108.7 slg 32.2 ft/s2 (1) ⇒

a(120 ft) =

 1 0 − (88 ft/s)2 2

a = −32.26 ft/s2 The force acting on the body (Fb ) times the speed of the body gives us the power and the force acting on the body is simply the mass times the acceleration: P = Fb v = mav = (108.7 slg)(−32.26 ft/s2 )v = −(3507 slg· ft/s2 )v The easiest way to plot this versus displacement, as asked for, is to use our velocity/displacement relationships (32.26 ft/s2 )t2 v(t) = 88 ft/s − (32.26)t, x(t) = (88 ft/s)t − , 2 compute v and x from t = 0 to t = 2.727 s (the start and finish times of the braking maneuver) and then use v to compute P and plot versus x. Doing so in MATLAB leads to the following plot of P vs x

205

206

4.3.29 GOAL: Determine the power delivered to a spinning wheel. GIVEN: System configuration, relevant masses, angular velocity of the wheel and dynamics coefficient of friction. DRAW:

FORMULATE EQUATIONS: We’ll use the power/speed relationship P = Fv where F is the force due to friction and v is the speed of the outer periphery of the wheel. SOLVE: The mass is slipping against the periphery of the wheel and therefore feels a frictional force F = µN = µmg cos θ (pointing in a clockwise direction) and a gravitational force mg sin θ pointing counter-clockwise. These forces have to balance so that the mass stays fixed in a particular orientation: µmg cos θ = mg sin θ tan θ = µ ⇒ θ = 11.3◦ We now know the steady-state position of the mass with respect to the wheel. As the FBD shows, the wheel has to supply power to counter the force that’s continually trying to rotate the wheel counter-clockwise. The power required is equal to this frictional force F multiplied by the slip speed: P = (µmg cos θ)vslip = 0.2(0.08 kg)(9.81 m/s2 )(0.981)(15 rad/s)(0.1 m) P = 0.231 W

207