O

A

datum O

ME 274 – Spring 2018 Final Examination – Problem 1 1.FBD-above2.KineticsSincenonon-conservativeworkisdoneonthesystem,energyisconserved:

(1) T1 +V1 = T2 +V2 ⇒ 0+ mgH = 1

2mvA2

2 + 12

mvO22 + 1

2IOω2

2 + 0 ; IO = 12

mR2

Sincenoforcesactonthesysteminthex-direction,linearmomentuminx-directionisconserved:(2) 0 = mvA2 + mvO2 ⇒ vA2 = −vO2

3.Kinematics

(3) !vO2 = vO2i = !vA2 +

!ω2 ×

!rO/ A = −vO2i + ω2k( )× Rj( ) = − vO2 + Rω2( ) i ⇒ vO2 = − R2ω2

4.SolveCombining(1)-(3):

mgH = 2( ) 1

2mvO2

2 + 12

12

mR2⎛⎝⎜

⎞⎠⎟ω2

2 = m R2ω2

⎛⎝⎜

⎞⎠⎟

2

+ 14

mR2ω22 = 1

2mR2ω2

2 ⇒!ω2 = − 2gH

R2 k

ME 274: Basic Mechanics IISpring 2018

April 30, 2018

Problem 1

Given: A homogeneous disk of mass m and outer radius R is able to roll without slipping on thecurved upper surface of a cart A. Cart A (of mass m) is able to move along a smooth, horizontalsurface. The system is released from rest at Position 1. At Position 2, the center O of the diskhas dropped through a vertical distance of H, and the disk is rolling on a horizontal portion of theupper surface of cart A.

Find: For this problem:

(a) Draw a single free body diagram (FBD) of the system made up of the disk and the cart.(b) Explain why mechanical energy is conserved for the system shown in your FBD.(c) Explain why linear momentum in the x-direction is conserved for the system shown in your

FBD.(d) Determine the angular velocity of the disk at Position 2. Write your answer as a vector.

smooth m

Posi%on1

C

C

O

A

H

R

no slip

m

Posi%on2

x

y

O

Final Examination Page 2 of 16

ME 274: Basic Mechanics IISpring 2018

April 30, 2018

Problem 2

Given: A homogeneous disk of mass m and outer radius R is able to roll without slipping on arough horizontal surface. A bar A (of mass m) is in a no-slip contact with the top of the disk withthe other end of A being supported by a smooth roller. A spring of sti↵ness k is connected betweenthe right end of bar A and ground. A second spring (of sti↵ness 2k) is connected between thecenter O of the disk and a movable base B. Base B is given a prescribed motion of xB(t) = b sin ⌦t.Let x represent the motion of the bar A, where x = xB = 0 when the springs are unstretched.

Find: For this problem:

(a) Draw individual free body diagrams of the disk and of the bar A.(b) Derive the di↵erential equation of motion (EOM) for the system in terms of the coordinate x.(c) What is the natural frequency !n of the system?(d) Determine the particular solution xP (t) of your EOM.(e) Based on your particular solution above, is the motion of the bar in phase or 180� out of phase

with the motion of the base B when ⌦ =p

k/m?

smooth O

no slip

m

m

R

C

k

2k

A

B no slip

x+

xB(t)

Final Examination Page 5 of 16

ME 274 – Spring 2018 Final Examination – Problem 2 1.FBDs-shown

2.Newton-EulerBar: Fx∑ = −kx + fA = m!!x

Disk:

MC∑ = −2k Rθ − xB( )R − fA 2R( ) = IC!!θ ; IC = IO + mR2 = 1

2mR2 + mR2 = 3

2mR2

Therefore:

(1) fA = m!!x + kx = − 3

4mR!!θ − k Rθ − xB( )

3.KinematicsSinceCistheICforthedisk:(2) !x = 2R !θ ⇒ !!θ = !!x / 2R AND θ = x / 2R

4.EOMCombining(1)and(2):

m!!x + kx = − 3

4mR

!!x2R

⎛⎝⎜

⎞⎠⎟− k R

x2R

− xB⎛⎝⎜

⎞⎠⎟

⇒ 118

m!!x + 32

kx = kb sinΩt

DivideEOMbythecoefficientof !!x gives:

!!x + 12

11km

x = 811

km

b sinΩt ⇒ !!x +ωn2x = F0 sinΩt ; F0 =

811

km

b

with: ωn =

1211

km

kx

θ +

fC

fA

fA

k Rθ − xB( )

Particularsolution

xP t( ) = AcosΩt + BsinΩt

!!xP t( ) = −Ω2 AcosΩt −Ω2BsinΩt

SubstitutingintheEOMgives:

−Ω2 +ωn2( )AcosΩt + −Ω2 +ωn

2( )BsinΩt = F0sinΩt ⇒

cosΩt : A = 0

sinΩt : −Ω2 +ωn2( )B = F0 ⇒ B =

8 / 11( ) k / m( )−Ω2 +ωn

2 b

Since Ω = k / m <ωn ⇒ B > 0 ⇒ particularsolutionresponseisINPHASEwiththebasemotion.

ME 274 – Spring 2018 Final Examination – Problem 3A

EOM:

!!x − 13!x + 36x = 0 ⇒ 2ζωn = −1/ 3 ⇒ ζ < 0 ⇒

x t( ) = eζ ωnt C cosωdt + S sinωdt( ) ⇒ exponentially GROWING oscillations

ME 274: Basic Mechanics IISpring 2018

April 30, 2018

Problem 3 - PART A (2 points)

Given: The following equation of motion (EOM) has been derived for a single-degree-of-freedomsystem: 6x � 2x + 216x = 0.

Find: Circle the figure below that most accurately represents the response that is described bythe above EOM.

time, t

resp

onse

, x(t)

time, t

resp

onse

, x(t)

time, t

resp

onse

, x(t)

Final Examination Page 8 of 16

ME 274 – Spring 2018 Final Examination – Problem 3B

SystemA:

Fx∑ = −kx = m!!x ⇒ m!!x + kx = 0 ⇒ !!x + k

mx = 0 ⇒ ωnA = k

m

SystemB:

MC∑ = −kx R( ) = IC

!!θ = 32

mR2⎛⎝⎜

⎞⎠⎟!!xR

⎛⎝⎜

⎞⎠⎟⇒ 3

2m!!x + kx = 0 ⇒ !!x + k

3mx = 0 ⇒ ωnB = k

3m

Therefore, ωnA >ωnB .

ME 274: Basic Mechanics IISpring 2018

April 30, 2018

Problem 3 - PART B (2 points)

Given: Consider Systems A and B shown below. System A is made up of a spring and block withthe block moving in pure translation along a smooth horizontal surface. System B is made up ofa spring and a homogeneous disk of mass m and outer radius R, with the center of the disk at Oand the disk rolling without slipping on a horizontal surface. Each system has the same mass mand same spring sti↵ness k. Let !nA and !nB represent the natural frequencies of Systems A andB, respectively.

Find: Circle the answer below that most accurately represents the natural frequencies for the twosystems:

(a) !nA < !nB

(b) !nA = !nB

(c) !nA > !nB

Chapter

6

Question C6.8

Consider Systems A and B shown below. System A is made up of a spring and block with theblock moving in pure translation along a smooth horizontal surface. System B is made up of aspring and a homogeneous disk of mass m and outer radius R, with the center of the disk at O andthe disk rolling without slipping on a horizontal surface. Each system has the same mass m andsame spring sti↵ness k. Let !nA and !nB represent the natural frequencies of Systems A and B,respectively. Circle the answer below that most accurately represents the natural frequencies forthe two systems:

(a) !nA > !nB

(b) !nA = !nB

(c) !nA < !nB

(d) More information is needed on the two systems in order to answer this question.

k c

m g

equilibrium position

unstretched position

z x

xst

A

k m

smooth

x

no slip

m k

O R

x

System A System B

Final Examination Page 9 of 13

kx kx

C f

ME 274 – Spring 2018 Final Examination – Problem 3C

m!!x + c !x + kx = 0 ⇒ !!x + c

m!x + k

mx = 0 ⇒ ωn =

km

& ζ = c2ωnm

= c

2 km

a)

ζ new = c

2 2k( ) 2m( )= 1

2ζ old

b)

ζ new = 2c

2 2k( )m= 2ζ old

ME 274: Basic Mechanics IISpring 2018

April 30, 2018

Problem 3 - PART C (4 points)

Given: Consider the free response for a damped, single-DOF system having the following di↵er-ential equation of motion: mx + cx + kx = 0. This system is known to have a damping ratio of ⇣= 0.1 and undamped natural frequency of !n = 10 rad/s.

Find: What is the new value for the damping ratio ⇣:

(a) the original value of m is doubled, the original value of k is doubled and the value of c isunchanged?

(b) The original value of k is doubled, the original value of c is doubled and the value of m isunchanged?

Final Examination Page 10 of 13

ME 274 – Spring 2018 Final Examination – Problem 3D

NotethatCistheICforthedrum.Therefore,astheforceFisapplied,thedrumwillrotateCWaboutC.Asaresult,thecontactpointAwillmovetotheLEFT.SincefrictionopposedtheslidingmotionofA,theslidingfrictionforce f = µkmgwillacttotheRIGHT.

ME 274: Basic Mechanics IISpring 2018

April 30, 2018

Problem 3 - PART D (2 points)

Given: A cable is wrapped around the inner radius of a stepped drum, with the other end of thecable attached to ground, as shown in the figure. A second cable is wrapped around the outerradius of the drum, with a force F applied to the free end of that cable. The drum is supportedby a horizontal surface on which the drum slips at contact point A as the drum moves. Neithercable slips on the drum as the system moves. Let µs and µk be the coe�cients of static and kineticfriction, respectively, between the drum and the supporting surface at A.

Find: Circle the answer below that most accurately describes the friction force f acting on thedrum at the contact point A.

(a) f = µsmg (to the left)(b) f = µkmg (to the left)(c) f = 0(d) f = µkmg (to the right)(e) f = µsmg (to the right)

no slip

2R R

F

C

slip

O

taut cable

A

Final Examination Page 11 of 13

ME 274 – Spring 2018 Final Examination – Problem 3E

ForthesystemofthebarandparticleP,

!MO∑ =

!0 .Therefore:

!HO1 =

!HO2 ⇒ m!rP/O × !vP1 = m!rP/O × !vP2 ⇒

R02

eR⎛⎝⎜

⎞⎠⎟×

R02ω1eθ

⎛⎝⎜

⎞⎠⎟= R0eR( )× "ReR + R0ω2eθ( ) ⇒

R02

4ω1k = R0

2ω2k ⇒ ω2 =14ω1 INDEPENDENT of k( )

ME 274: Basic Mechanics IISpring 2018

April 30, 2018

Problem 3 - PART E (2 points)

Given: A particle P of mass m is free to slide on a smooth bar of negligible mass. The bar is freeto rotate in a horizontal plane about a vertical axis passing through end O of the bar. A spring ofsti↵ness k and unstretched length R0 is connected between P and O. The spring is compressed tohalf of its unstretched length and released when the bar has a rotational speed of !1. After release,P reaches a position when the spring is unstretched. At this position, the rotational speed of thebar is !2.

Find: Suppose now the experiment is repeated except the sti↵ness of the spring is doubled to avalue of 2k. As a result of this change, the value of !2 is now:

(a) decreased.(b) unchanged.(c) increased.

Question C4.13

A particle P is free to slide on a smooth, lightweight bar. The bar is free to rotate in a horizontalplane about a vertical axis passing through end O of the bar. A spring of sti↵ness k and unstretchedlength R0 is connected between P and O. The spring is compressed to half of its unstretched lengthand released when the bar has a rotational speed of !1. After release, P reaches a position whenthe spring is unstretched. At this position, the rotational speed of the bar is !2.

Suppose now the experiment is repeated except the sti↵ness of the spring is doubled to a value of2k. As a result of this change, the value of !2 is now:

(a) Decreased

(b) The same

(c) Increased

(d) More information is needed in order to answer this question.

ME 274 – Spring 2007 Name

Examination No. 2

PROBLEM NO. 3 Instructor

Part (d) – 5 points

A particle P is free to slide on a smooth, lightweight bar. The bar is free to rotate in a

horizontal plane about a vertical shaft passing through O. A spring of stiffness k and

unstretched length R0 is connected between P and O. The spring is compressed to half of

its unstretched length and released when the bar has a rotational speed of !1. After

release P reaches a position where the spring is unstretched. At this position, the

rotational speed of the bar is !2.

Suppose now that the experiment is repeated except the stiffness of the spring is doubled

to a value of 2k. As a result, the value of !2 is:

a) is decreased.

b) remains the same.

c) is increased.

d) more information is needed to answer this question.

Provide a justification for your answer.

P O

R0/2

HORIZONTAL PLANE

k

P

O

R0

k

!2

!1

Final Examination Page 12 of 13

ME 274 – Spring 2018 Final Examination – Problem 3E

!vA/ B( )rel

= !vA − !vB −!ω × !rA/ B = !vA/ B −

!ω × !rA/ B = vel. of A as seen by B .

Here !ω = ang. vel. of B =

!0 .Therefore:

!vA/ B( )rel

= !vA/ B

!vB/ A( )rel

= !vB − !vA −!ω × !rB/ A = !vB/ A −

!ω × !rB/ A = vel. of B as seen by A .

Here !ω = ang. vel. of A ≠

!0 .Therefore:

!vB/ A( )rel

≠ !vB/ A

ME 274: Basic Mechanics IISpring 2018

April 30, 2018

Problem 3 - PART F (2 points)

Given: Aircraft B is traveling along a straight path with a speed of vB. Aircraft A is travelingalong a circular path with a speed of vA.

Find: Circle the answer below that most accurately describes the observed velocities of A and B:

(a) ~vA/B = ~vA � ~vB is the velocity of A as seen by the pilot of B.(b) ~vB/A = ~vB � ~vA is the velocity of B as seen by the pilot of A.(c) Both (a) and (b).(d) Neither (a) nor (b).

ME274:BasicMechanicsIISpring2018

April30,2018

Problem3-PARTI(2points)

Given:AircraftAistravelingalongastraightpathwithaspeedofvA.AircraftBistravelingalongacircularpathofradiusRwithaspeedofvB.

Find:CircletheanswerbelowthatmostaccuratelydescribestheobservedvelocitiesofAandB:

(a)~vA/B=~vA�~vBisthevelocityofAasseenbythepilotofB.(b)~vB/A=~vB�~vAisthevelocityofBasseenbythepilotofA.(c)Both(a)and(b).(d)Neither(a)nor(b).

QuestionC1.10

AircraftAistravelingalongastraightpathwithaspeedofvA.AircraftBistravelingalongacircularpathofradiusRwithaspeedofvB.CircletheanswerbelowthatmostaccuratelydescribestheobservedvelocitiesofAandB:

(a)~vA/B=~vA�~vBisthevelocityofAasseenbythepilotofB

(b)~vB/A=~vB�~vAisthevelocityofBasseenbythepilotofA

(c)Both(a)and(b).

(d)Neither(a)nor(b).

!vB

!vA

B A

FinalExaminationPage12of12

!vA

!vB

B A

Final Examination Page 13 of 13