Free-body Diagram

List of 13 topics you might see on final exam (hmm…) z z z z z z z z z z z z z

x-y motion subject to constant acceleration Sliding friction, line tension Centripetal acceleration, UCM, gravity Work and energy, rotational and translational KE, Conservation of mechanical energy, potential energy, KE Center of Mass (multi body – several FBDs) Collisions, momentum conservation, elastic, inelastic (P cons) Rotational dynamics, moment of inertia Angular momentum, torque, spinning skater, work done by Statics, including forces and torques Simple harmonic motion (pendulum, spring Æ PE=½kx2) Waves (amplitude, k, omega, velocity, waves on rope, T) Buoyancy Physics 211: Review, Pg 1

Physics 211: Review Problems

¾ Dynamics (free body diagrams, net force) ¾ Centripetal acceleration (when do you use v2/R?) ¾ Work & Energy (kinetic, potential, external forces, …) ¾ Friction (static and dynamic) ¾ Center of mass motion and acceleration (Fext = MTACM) ¾ Momentum conservation & collisions (elastic and inelastic) ¾ Rotational dynamics (cons of angular momentum, torque) ¾ Simple Harmonic Motion (springs, pendula, omega…) ¾ Waves (wave number, angular freq, velocity, direction) ¾ Buoyancy

Physics 211: Review, Pg 2

Dynamics: z

Three blocks are connected by massless strings and frictionless pulleys as shown. The kinetic coefficient of friction between block C and the table is µ=0.5. The masses of blocks C and A are MC=20kg and MA=6kg. Block C accelerates to the right with a = 0.7m/s2. ¾ What are the tensions in the strings, TA and TB? ¾ What is MB. a MC TA MA

TB MB Physics 211: Review, Pg 3

Dynamics... z

First find TA: Consider F = ma for mass A: ¾ TA - MAg = MAa TA = MA(a + g) = 63N TA a MAg

Physics 211: Review, Pg 4

Dynamics... z

Next find TB: Consider F = ma for mass C: ¾ TB - TA - µMCg = MCa TB = TA + MC(a + µg) = 175N

a TA f = µMCg

MC N

TB MCg

Physics 211: Review, Pg 5

Dynamics... z

Finally find MB: Consider F = ma for mass B: ¾ MBg - TB = MBa ¾ MB(g - a) = TB MB =

TB =19 .2 kg g −a

TB a MBg

Physics 211: Review, Pg 6

Problem: Motion in a Circle

z

Tetherbal l A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v. ¾ What is the tension T in the string at the top of the rock’s trajectory?

v T R

Physics 211: Review, Pg 7

Motion in a Circle... z z z

Draw a Free Body Diagram (pick y-direction to be down): We will use FNET = ma (surprise) y First find FNET in y direction: FNET = mg +T

mg T

Physics 211: Review, Pg 8

Motion in a Circle... FNET = mg +T z z

v

Acceleration in y direction is known: It’s centripetal… ma = mv2 / R

y

mg T

F = ma

mg + T = mv2 / R

R T = mv2 / R - mg

Physics 211: Review, Pg 9

Motion in a Circle...

z

Bucke t What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp? ¾ i.e. find v such that T = 0.

v

mv2 / R = mg + T v2 / R = g

mg T= 0

v = Rg z

Notice that this does not depend on m.

R

Physics 211: Review, Pg 10

Work & Energy: z

A cart starts up a hill with initial speed v0. The body of the cart has mass m, and each of its four wheels has mass m (and radius r ). The wheels are uniform disks, and roll without slipping. Give answers in terms of m, v0 and g only. ¾ What is the initial kinetic energy K of the cart? ¾ How high up the hill h does the CM of the cart rise?

v0

h

Physics 211: Review, Pg 11

Work & Energy... z

The kinetic energy has both translational and rotational contributions: 1 1

KE INITIAL =

z

2

(5m )v02 + 4 ⋅

2

Iω 2

v0 1 2 ω = For a disk* I = mr and since the wheels don’t slip r 2

KE INITIAL

5 2 1 1 2 v02 = mv0 + 4 ⋅ ⋅ mr ⋅ 2 2 2 2 r v0

KE INITIAL =

7 2 mv0 2

This is only simple because both chassis and all the wheels had same mass!

*about its center Physics 211: Review, Pg 12

Work & Energy... z

Since energy is conserved, ∆K = -∆U: ∆K = −

7 mv 02 2

∆U = 5 mgh

7 v 02 h= 10 g

v=0 v0

h

Physics 211: Review, Pg 13

Center of Mass: z

Three uniform spherical masses are shown. Two are moving with the indicated velocities, and the third (at the origin) is stationary. ¾ What is the location and velocity of the center or mass? ¾ What is the total momentum of the system?

y(m)

v2 = (10,20)m/s

5 4 m3=5kg 3 2 m2=3kg v3 = (15,-5)m/s 1 x (m) m1=2kg 1 2 3 4 5 Physics 211: Review, Pg 14

Center of Mass... z

Rcm =

Location of CM:

X cm = Ycm =

1 M tot

∑ mi x i =

1 M tot

∑ mi y i

i

i

=

1 ∑ m i ri Mtot i

1 ( 3 kg ⋅ 2 m + 5 kg ⋅ 5 m ) = 31. m 10 kg

1 ( 3 kg ⋅ 3 m + 5 kg ⋅ 2 m ) = 1.9 m 10 kg y(m)

v2 = (10,20)m/s

5 4 m3=5kg 3 2 m2=3kg v3 = (15,-5)m/s 1 x (m) m1=2kg 1 2 3 4 5 Physics 211: Review, Pg 15

Center of Mass... z

Velocity of CM: Vcm =

Vx ,cm = V y ,cm =

1 M tot

∑ m i v x ,i =

1 M tot

∑ m i v y ,i

i

i

=

1 ∑ mi v i Mtot i

1 ( 3 kg ⋅10 m s + 5 kg ⋅15 m s ) = 10 .5 m s 10 kg 1 ( 3 kg ⋅ 20 m s − 5 kg ⋅ 5 m s ) = 3 .5 m s 10 kg y(m)

v2 = (10,20)m/s

5 4 m3=5kg 3 2 m2=3kg v3 = (15,-5)m/s 1 x (m) m1=2kg 1 2 3 4 5 Physics 211: Review, Pg 16

Center of Mass... z

Momentum of CM:

(

Pcm = M tot Vcm

)

Pcm = M totVx ,cm , M totVy ,cm = (10 kg ⋅10 .5 m s ,10 kg ⋅ 3 .5 m s ) Pcm = (105 ,35 )kg m s y(m)

v2 = (10,20)m/s

5 4 m3=5kg 3 2 m2=3kg v3 = (15,-5)m/s 1 x (m) m1=2kg 1 2 3 4 5 Physics 211: Review, Pg 17

Momentum conservation & collisions: z

Two particles collide elastically as shown below. The small particle has mass m = 1kg and is initially moving in the x direction. The final velocity of m is 5m/s in the y direction, and the final momentum of M is 26.57o below horizontal. ¾ Find vi and P. ¾ What is the kinetic energy of M after the collision? ¾ Determine the mass M. vf m y m

vi before

M

θ = 26.57o after

M

P

Physics 211: Review, Pg 18

x

Momentum conservation & collisions... z

Conserve momentum in the x and y directions: ¾ y: mv f = P sin θ

mv f (1kg ) (5 m s ) P= = = 11.2 kg m s sin θ sin( 26 .57 )

¾ x: mv i = P cos θ

vi =

P cos θ = 10 m s m

vf m m

vi before

M

y θ = 26.57o

after

M

P

Physics 211: Review, Pg 19

x

Momentum conservation & collisions... z

Conserve energy (its elastic): 1 1 1 mv i2 = mv f2 + K M K M = m v i2 − v f2 = 37 .5 J 2 2 2

(

and since K M

P2 = 2M

)

P2 M= = 1.67 kg 2K M

vf m m

vi before

y

M x after

M

P

Physics 211: Review, Pg 20

Rotational Dynamics: z

A circular cylinder of radius R = .2m, which can rotate freely on a horizontal, frictionless axle, has a light rope wound around it. The rope passes over a smooth peg and is attached to a hanging mass of mass m = 1kg. As m falls, the rope unwinds from the cylinder, whose moment of inertia Iz is not initially known. The mass accelerates downward with a = 0.4m/s2. ¾ What is the angular acceleration α of the cylinder about the axle? T ¾ What is the tension T in the rope? ¾ What is the moment of inertia Iz R of the cylinder about the axle? Iz ¾ What is the angular speed of the m cylinder after m falls a distance d = 10m ? a Physics 211: Review, Pg 21

Rotational Dynamics... z

What is the angular acceleration α of the cylinder about the axle? ¾ Since the string does not slip on the wheel, the magnitude of the acceleration of a point on the rim of the wheen is a = Rα.

α=

2

a .4 m s = = 2 rad s 2 .2 m R

T

α R Iz

m a Physics 211: Review, Pg 22

Rotational Dynamics... z

What is the tension T in the rope? ¾ Using F = ma for the hanging mass in the “down” direction:

T m

mg - T = ma

mg

¾ T = m(g - a) = 1kg (9.8 m/s2 - 0.4 m/s2) = 9.4 N

Physics 211: Review, Pg 23

Rotational Dynamics... z

What is the moment of inertia Iz of the cylinder about the axle? ¾ Using τ = Izα for the cylinder: TR = Iz α

Iz =

Iz =

TR α

( 9 .4N ) (.2 m ) 2 rad s 2

α

T R Iz

= 0 .94 kgm 2

Physics 211: Review, Pg 24

Rotational Dynamics... z

What is the angular speed of the cylinder after m falls a distance d = 10m ? ¾ Using energy conservation, ∆K = -∆U: 1 1 Iz ω f2 + mv f2 = mgd ωf 2 2 Iz

but vf = Rωf 1 1 Iz ω f2 + mR 2 ω f2 = mgd 2 2 ωf =

2 mgd . rad s = 1414 2 Iz + mR

d m vf Physics 211: Review, Pg 25

Angular Momentum : z

A student wishes to close a door withour getting out of bed. She throws a shoe of mass m = 600gm which hits the door exactly in the center. The shoe has horizontal speed v0 = 8m/s just before hitting the door and it bounces back with horizontal speed v1 = 4m/s. The door has width w = 1m. Just after the collision, the door turns with angular velocity ωz = 0.75 rad/sec. ¾ What is the moment of inertia of the door about the hinge? hinge

v0

w=1m

ωz

v1 top views Physics 211: Review, Pg 26

Angular Momentum ... z

Since there are no external torques, angular momentum about the hinge (z-axis) is conserved:

Lz ,before = mv 0

w 2

Lz ,after = I ω z − mv1

= m (v 0 + v1 )w I= 2 ωz

w 2

= 4.8 kgm 2 ωz

v0

w=1m

v1

Physics 211: Review, Pg 27

SHM and Springs

d 2s 2 = − ω s 2 dt

Force:

ω=

k m

k s Solution: s = A cos(ωt + φ)

k

m 0

0 m

s Physics 211: Review, Pg 28

Velocity and Acceleration with Springs Position: x(t) = A cos(ωt + φ) Velocity: v(t) = -ωA sin(ωt + φ) Acceleration: a(t) = -ω2A cos(ωt + φ) xMAX = A vMAX = ωA aMAX = ω2A k

by taking derivatives, since:

v( t ) =

dx ( t ) dt

a( t ) =

dv ( t ) dt

m U ( x) = 12 kx 2

0

x Physics 211: Review, Pg 29

Simple Harmonic Motion: z

A physical pendulum consists of two equal uniform disks attached rigidly together as shown. Each disk has mass M and radius R. The pendulum rotates freely about a pivot at the center of one disk. Give answers in terms of M, R and g. ¾ Find the moment of inertia I about the pivot. ¾ Find the period of oscillation T for small displacements. pivot R M

Physics 211: Review, Pg 30

Simple Harmonic Motion... z

The moment of inertia will have contributions from both disks. I=

1 1 MR 2 + MR 2 + M ( 2 R ) 2 2 2

top disk

bottom disk

pivot R M 2R

I = 5 MR 2

R

M

Physics 211: Review, Pg 31

Simple Harmonic Motion... z

The angular frequency MgD of a physical pendulum is: ω = I ¾ Where D is the distance from the pivot to the CM, ¾ I is the moment of inertia about the pivot, ¾ and M is the total mass

2 MgR 2g = so in this case: ω = 5R 5 MR 2 T =

pivot R

CM

2π 5R = 2π ω 2g

2Mg Physics 211: Review, Pg 32

Waves: z

A wave given by y = 1.5 cos(0.1x - 560t ), with x and y in cm and t in seconds. The wire tension is 28 N. Find: ¾ the amplitude, ¾ the wavelength, ¾ the period, ¾ the wave speed, ¾ the power carried by the wave.

y x

Physics 211: Review, Pg 33

Waves... y

λ A x

y = 1.5 cos(0.1x - 560t )

Compare this to the generic form: y = A cos(kx − ω t ) A = 1.5 cm = 0.015 m k = 0.1 cm-1 = 10 m-1 ω = 560 rad/s

Physics 211: Review, Pg 34

y

Waves...

λ A x

k=

2π λ

y = .015 cos(10x - 560t ) in SI units

so λ = 0 .628 m

ω = 560 rad/s

v = λf =

ω k

T =

2π = .011 sec ω

v = 56 m/s

Physics 211: Review, Pg 35

y

Waves...

λ A x

Finally P =

v =

F µ

1 µv ω 2 A 2 2

µ =

y = .015 cos(10x - 560t )

So we need to figure out µ

28 N F = = .0 0 8 9 k g m 2 2 v (5 6 m s )

Physics 211: Review, Pg 36

y

Waves...

λ A x

So

P =

y = .015 cos(10x - 560t )

1 µv ω 2 A 2 2

But we know that A = 0.015 m v = 56 m/s ω = 560 rad/s µ = 0.0089 kg/m

P = 1 7 .6 J / s

Physics 211: Review, Pg 37

Buoyancy A plastic cube with length 0.3 m on each side is floating in a pool of water (ρ = 1000 kg/m3). A cylinder with volume VC = 0.005 m3 is supported by a string attached to the upper cube. The tension T in the string attaching the two objects is 24.5 N. The system is at equilibrium with the cube submerged a distance h = 0.2 m.

h

0.3m T = 24.5N

• What is the mass of the cylinder • What is the mass of the cube • How would h change if the cylinder were placed on top of the cube? Physics 211: Review, Pg 38

Buoyancy FB

T = 24.5N

mg

FB = VcρWg

= (0.005m3)(1000kg/m3)(9.81m/s2) = 49N

mg = T + FB

m = (T + FB)/g = (24.5N+49N)/(9.81m/s2) = 7.5 kg

Physics 211: Review, Pg 39

Buoyancy VSUB=(0.3)(0.3)(0.2)m3 = 0.018m3 FB = VSUBρWg = (0.018m3)(1000kg/m3)(9.81m/s2) = 177N 0.3m

0.2 Mg

Mg + T = FB

T = 24.5N

M = (FB - T)/g = (177N – 24.5N)/(9.81m/s2) = 15.5 kg Physics 211: Review, Pg 40

Buoyancy

h

hnew

hnew > h since the submerged volume has to be the same

Physics 211: Review, Pg 41

Thanks for a great semester! Good luck on the final !! Have a really great summer! Welcome back next fall!!! Physics 211: Review, Pg 42

Kinematics: z

Bill stands on the roof of his house, a distance d=13m above the ground, operating a pitching machine. At t = 0 he launches a ball with initial speed v0 at an angle θ above horizontal. The ball hits the ground at the base of his friend Ted’s house, a distance D = 80m away, at exactly tf = 4s. ¾ What are v0 and θ ? ¾ What is the maximum height H reached by the ball ? ¾ How fast vf is the ball moving when it hits the ground ? v0 Bill

θ d=13m

vf

Ted

D=80m Physics 211: Review, Pg 43

Kinematics... z z

The distance traveled in the x direction is: D = v 0 cos θ ⋅ t f (a) The y position of the ball is given by: 1 y = y 0 + v y 0 t − gt 2 2 1 2 − d = v ⋅ t − gt f sin θ which at t = tf becomes: 0 f 2 1 v 0 sin θ ⋅ t f = gt f2 − d (b) 2

dividing (b) by (a):

gt f2 − 2 d tan θ = 2D

Physics 211: Review, Pg 44

Kinematics... z

gt f2 − 2 d Plug the given numbers into tan θ = 2D θ = 39 .3 o

z

Now put this back into D = v 0 cos θ ⋅ t f and solve for v0: v0 =

D cos θ ⋅ t f

v 0 = 25 .8 m s

Physics 211: Review, Pg 45

Kinematics... z

The time at the top of the trajectory tt can be found by solving vy = vy0 - gtt = 0

z

Plug this into the y-position equation:

v 0 sin θ tt = = g g v y0

1 2 (v 0 sin θ) h = v 0 sin θ ⋅ t t − gt t = = 13 .6 m 2 2g 2

So: H = d + h = 26.6m tt v0 Bill

θ d=13m

h H

Ted

Physics 211: Review, Pg 46

Kinematics... z

The y-component of the velocity at t = tf can be found by v yf = v 0 sin θ − gt f = −22 .9 m s plugging into vyf = vy0 - gtf

z

The x-component of velocity is constant: v x = v 0 cos θ = 20 m s

z

The final speed is then: v f = v xf2 + v yf2 = 30 .4 m s (could also find this using energy conservation) v0 Bill

θ

tf vf

Ted

Physics 211: Review, Pg 47