Final Soln

ME 274 – Fall 2008 Final Examination PROBLEM NO. 1

SOLUTION

Given:

A solid drum having a mass of 2m and centroidal radius of gyration kO is pinned to ground at its center of mass O. A cable that is wrapped around the outer radius (2R) of the drum supports block A that has a mass of 3m. A second cable that is wrapped around the inner radius (R) of the drum supports block B that has a mass of m. Assume that the cables do not slip on the drum as the system moves. The system is released from rest.

Find:

Determine the initial angular acceleration of the drum on release. Write your answer as a vector. As always, clearly show the four steps in your solution.

Oy 2m H

2R

Ox

O

O

E

g 2mg

R TA

TB

B

A

m

B

A

3m mg 3mg

1. FBD’s: shown to the right 2. Newton-Euler: Drum : ! M O = TA ( 2R ) " TB ( R ) = IO# = ( 2m ) kO2 #

! Fy = TA " 3mg = 3maA # TA = 3m ( g + aA ) B : ! Fy = TB " mg = maB # TB = m ( g + aB ) A:

Therefore, (1) : 6mR ( g + aA ) ! mR ( g + aB ) = 2mkO2 "

3. Kinematics: a H = aO + ! " r H /O # $ 2 r H /O = 0 + (! k ) " ( #2Ri ) # 0 = #2R! j

a E = aO + ! " r E /O # $ 2 r E /O = 0 + (! k ) " ( Ri ) # 0 = R! j

( 2 ) : aA = aH = !2R" ( 3) : aB = aE = R" 4. Solve: Combining (1)-(3) gives

6R ( g ! 2R" ) ! R ( g + R" ) = 2kO2 " # " = Therefore,

" % 5gR ! =$ 2 k 2' # 2kO + 13R &

5gR 2kO2 + 13R 2

ME 274 – Fall 2008 Final Examination PROBLEM NO. 2

SOLUTION

Given:

A homogeneous disk having a mass of m and radius R rolls without slipping on a rough, horizontal surface. A spring of stiffness k connects the center of the disk O to ground. A second spring, of stiffness 4k, is attached between O and a moving base B. Base B is given a prescribed motion of xB(t) = b sin!t. The coordinate x describes the location of the center of the disk O. The two springs are unstretched when x = xB = 0.

Find:

For this problem, a) DERIVE the differential equation of motion (EOM) of the system. As always, clearly show your four steps in the analysis. b) Write down the natural frequency for the system. c) DERIVE the particular solution xP(t) for the system. Use the following parameter values in your final answer: m = 5 kg, k = 600 N/m, R = 0.3 meters, b = 0.1 meters and ! = 40 rad/sec. d) Does the disk move in phase or out of phase with the base?

xB(t)

x

m 4k

k

O R

B

no slip

1. FBD: shown to the right 2. Newton-Euler:

! M C = 4k ( x " xB ) R + ( kx ) R = IC#!! 3. Kinematics

; I C = IO + mR 2 =

3 mR 2 2

mg O kx

4k(x – xB) C

f N

aO = aC + ! " r O/C # $! 2 r O/C

( )

(

( )

)

!! x i = aC j + ($!! k ) " R j # $! 2 R j = ( #R$!!) i + aC # R$! 2 j i : !! x = #R$!!

%

%

$!! = #!! x/R

4. EOM: Combining above gives

4k ( x ! x B ) + kx =

3 x% 3 " !! mR $ ! ' = ! m!! x # R& 2 2

(

3 m!! x + 5kx = 4kx B ( t ) = 4kb sin) t 2 5k 10k = Natural frequency: ! n = 3m / 2 3m Particular solution: xP ( t ) = A sin! t + B cos! t Substitute into the EOM and group together like terms: # 3 & # 3 & 2 2 %$ ! m" + 5k (' A sin" t + %$ ! m" + 5k (' B cos" t = 4kb sin" t 2 2 # 3 & cos" t : % ! m" 2 + 5k ( B = 0 $ 2 ' # 3 & sin" t : % ! m" 2 + 5k ( A = 4kb $ 2 ' A=

4kb 3 ! m" 2 + 5k 2

=

)

)

B=0

)

4 ( 0.1) / 5 4b / 5 = = ! 0.0267 meters 3m 2 3) ( 5 ) ( 2 1! " 1! ( 40 ) 10k (10 ) ( 600 )

Therefore, xP ( t ) = ! 0.0267 sin40t Since A < 0, response is out of phase with the base motion.

ME 274 – Fall 2008 Final Examination PROBLEM NO. 3

SOLUTION

Given:

Particles A and B (having masses of mA = mB = 5 kg) are interconnected by the cable-pulley system shown in the figure. Both particles are constrained to vertical motion with particle A able to slide on a smooth vertical rod. The system is released at sA = 0 with A traveling DOWNWARD with a speed of 15 m/sec. Assume the pulleys to be small, massless and frictionless, and that the cable does not slip on the pulleys.

Find:

You are asked here to find the speed of particle A when A has reached the position of sA = 1.5 meters. As always, clearly indicate the four steps of your solution. 2m

O sA

Oy

T

O

sB

A g smooth rod

Ox N

A

B mAg

B mBg

FBD: shown on previous page. All forces doing work are conservative and will be included in the potential energy. Note that T does no work since it acts at a stationary point

Work-energy: 1 2 1 2 T1 = mvA1 + mvB1 2 2 V1 = 0 ; datum lines at initial positions of A and B

1 2 1 2 mvA2 + mvB2 2 2 V2 = !mgs A + mghB ( nc ) U1"2 = 0 T1 =

Therefore, ( nc ) T1 + V1 + U1!2 = T2 + V2

"

2 2 2 2 vA1 + vB1 = vA2 + vB2 + 2g ( #s A + hB )

Kinematics

2s B + s A + s A2 + 2 2 = const.

!

" sA % 1" sA % ' s!A = 0 ' vA 2 s!B + $1 + ! vB = $1 + 2 2 2$ $ ' ' s + 4 s + 4 A A # & # & 1 2 Also, !s B = # s A2 + 2 2 + s A2 " 2 & = hB % (' 2$ At position 1: vB1 =

1! 0 $ 1 + vA1 = 7.5m / sec 2 #" 0 + 4 &%

At position 2: $ 1! 1.5 1 2 2 vB2 = #1 + & vA2 = 0.8vA2 and hB = !#1.5 + 2 + 1.5 ' 2 $& = 1 m 2 " % 2 #" 2 1.5 + 4 &% Solve:

vA2 =

2 2 vA1 + vB1 + 2g ( s A ! hB )

1 + 0.8 2

=

(15 )2 + ( 7.5 )2 + ( 2 ) ( 9.806 ) (1.5 ! 1) = 13.3 m / sec 1 + 0.64

ME 274 – Fall 2008 Final Examination PROBLEM NO. 4

SOLUTION

Part (a) – 4 points

The mechanism shown below is made up of 5 moveable links: links AB, BD, DF, EG and HG. At the position shown, link AB is rotating clockwise (CW). Circle the answers below that most accurately describe the rotation of the other four links: Link BD:

!BD = CW

!BD = 0

!BD = CCW

Link DF:

!DF = CW

!DF = 0

!DF = CCW

Link EG:

!EG = CW

!EG = 0

!EG = CCW

Link HG:

!HG = CW

!HG = 0

!HG = CCW

HINT: Determine the locations of the instant centers for links BD and EG.

IEG

G

H

! EG ! HG

E

B

D

! AB A

F

! BD IBD

! DF

ME 274 – Fall 2008 Final Examination PROBLEM NO. 4

SOLUTION

Part (b) – 4 points

For the position shown (R = 3 ft and " = 2 radians), the velocity and acceleration of point P are known to be:

v P = ( 3e R ! 4 e" ) ft / sec

eR

a P = ( !20 e R ) ft / sec 2

P

respectively. Determine the rate of change of speed for P. e" !v $ ! 3e ' 4 e( $ v! = a P i et = a P i # P & = ( '20 e R ) i # R = '12 ft / sec 2 & " % 5 " vP %

"

R

O

ME 274 – Fall 2008 Final Examination PROBLEM NO. 4

SOLUTION

Part (c) – 4 points

Particle P slides within a smooth tube that is rotating about a vertical axis with a constant CCW rate of !. If R! < 0 (that is, P is sliding toward O), on which surface does P contact the tube: on surface A or surface B? Provide a justification for your answer. HINT: Draw an FBD of P and use Newton’s second law.

e"

!

surface A

O

R

P

eR

surface B

HORIZONTAL PLANE

" F! = N = m ( R#! + 2 R!# ) = 2mR!# Since ! > 0 (CCW) and R! < 0 (given), then N < 0. Therefore, P is in contact with surface A.

e"

eR N

ME 274 – Fall 2008 Final Examination PROBLEM NO. 4

SOLUTION

Part (d) – 4 points

Prior to impact, block B is moving to the left with a speed of vB1 and cart A is stationary (vA1 = 0). After impact block B is stationary (vB2 = 0) and block A moves to the left with a speed of vA2. What is the numerical value for the coefficient of restitution e for this impact? Consider all surfaces to be smooth. vB1

vA1 = 0

y

A B

3m

m

Position 1 (before impact)

! Fx = 0

"

mvB1 = 3mvA2

e=

3mvA1 + mvB1 = 3mvA2 + mvB2 "

vA2 1 = vB1 3

vA2 ! vB2 vA2 1 = = vB1 ! vA1 vB1 3

"

x

ME 274 – Fall 2008 Final Examination PROBLEM NO. 4

SOLUTION

Part (e) – 4 points

A homogeneous, thin bar of mass m and length L is released from rest with end A in contact with a smooth horizontal surface. Make an accurate sketch indicating the DIRECTION of the acceleration of the center of mass G immediately after release. HINT: Consider the FBD of the bar on release.

mg g

G

y

G x

60°

smooth

A

! Fx = 0

"

N

aGx = 0

Therefore, the acceleration of G is straight down (negative y-direction)

A

ME 274 – Fall 2008 Final Examination PROBLEM NO. 5

SOLUTION

Part (a) – 4 points

In lecture, the work-energy equation for a particle ( T2 = T1 + U1!2 ) was derived from Newton’s second law for a particle ( R = ma ). According to your instructor, what are the TWO critical steps in this derivation? • •

Projection of Newton’s second law onto et dv dv ds dv Using chain rule to produce: v! = = =v dt ds dt ds

R et

en

P

path of P

ME 274 – Fall 2008 Final Examination PROBLEM NO. 5

SOLUTION

Part (b) – 4 points

The positions of particles A and B are to be described by the coordinates xA and xB, respectively, defined below. All springs being unstretched when xA = xB = 0. In the FBD’s of A and B below, show the magnitude and direction of all spring forces in terms of k, xA and xB. k kxB k

k

B

A

xA

xB

k kxA

A

B k(xB- xA)

FBD’s of particles A and B

kxB

ME 274 – Fall 2008 Final Examination PROBLEM NO. 5

SOLUTION

Part (c) – 4 points

1 3 1 x ! sin (" x ) 3 2" (with x and y given in feet). When x = 1 ft, it is known that x! = !6 ft / sec = constant . At this position, what is !! y?

The path of a particle P moving in the x-y plane is given by: y ( x ) =

dy dy = x! ; chain rule dt dx 1 # & = % x 2 ! cos (" x ) ( x! 2 $ '

d2y dt

2

=

d # 2 1 1 & # & x ! cos (" x ) ( x! + % x 2 ! cos (" x ) ( !! x % dt $ 2 2 ' $ '

=

d # 2 1 & x ! cos (" x ) ( x! 2 % dx $ 2 '

;

product rule

; chain rule

" " # & # & 2 = % 2x + sin (" x ) ( x! 2 = % 2 (1) + sin (" ) ( ( !6 ) = 72 ft / sec 2 2 2 $ ' $ '

ME 274 – Fall 2008 Final Examination PROBLEM NO. 5

SOLUTION

Part (d) – 4 points

A wheel having an outer radius of R rolls without slipping with the center of the wheel O having a constant speed of vO to the right. Determine the acceleration vector for the contact point C. [You are asked to DERIVE this vector – no credit will be given for writing down the answer from memory.]

y vO

O

x

R C

no slip

aC = aO + ! " r C /O # $ 2 r C /O

(

= 0 + 0 # $ 2 #R j

)

= R$ 2 j 2

v2 %v ( = R' O * j = O j & R) R

Note that this is NOT a centripetal component of acceleration term since it points in the et direction for the path of C.

ME 274 – Fall 2008 Final Examination PROBLEM NO. 5

SOLUTION

Part (e) – 4 points

Consider the homogeneous solid wheel (of mass m, outer radius R and center A) and thin ring (of mass m, outer radius R and center B) shown below. The wheel and ring are both released from rest from the same height on identical rough inclines and roll down the inclines without slipping. After each has rolled down through a vertical height of h, how would you describe the relative sizes of the speeds of points A and B: a) vA > vB b) vA = vB c) vA < vB Justify your answer. m

m

A

B

R

R C

no slip

h

no slip

A

B DATUM

( nc ) T1 + V1 + U1!2 = T2 + V2

For solid wheel: I C =

"

mgh =

1 1 $ v' I C# 2 = I C & ) 2 2 % R(

1 3 mR 2 + mR 2 = mR 2 ! vA = 2 2

For ring: I C = mR 2 + mR 2 = 2mR 2 ! vB = mgh

!

2

"

4mgh 3

vB < vA

v=

2mgh IC / R2

C