me 274 – spring 2008 solution final examination ... · pdf file4 me 274 – spring...

1

ME 274 – Spring 2008 SOLUTION

Final Examination - Wednesday

PROBLEM NO. 1

Given: Disk P (having a mass of m = 2 kg) is constrained to move on a smooth,

horizontal surface. An inextensible cord is attached to P with the cord passing

through a hole in the surface at point O. A particle having a mass of M = 8 kg

is attached to the other end of the

cord, as shown below. At position 1

where R = 2 meters, P has a velocity

that is perpendicular to OP with v1 =

3 meters/sec. At position 2 where R =

1.5 meters, the velocity of P has both

eR and e! components.

Find: For position 2 of P,

a) find the value of !! = d! / dt .

b) find the value of !R = dR /dt .

c) write down the velocity vector for P in terms of its polar coordinates.

FBD of P – shown

Angular momentum

H

O( )1= mr

P /O! v

P1= m R

1e

R( ) ! v1e"( ) = m R

1v

1k

H

O( )2= mr

P /O! v

P2= m R

2e

R( ) ! !R2e

R+ R

2!"2e"( ) = m R

2

2 !"2

k

MO! = 0 " H

O( )1= H

O( )2

" m R1v

1= m R

2

2 !#2

" !#2=

R1v

1

R2

2=

2( ) 3( )

1.5( )2= 2.67 rad / sec

Work-energy (for P and B together)

T

1+V

1+U

1!2

nc( )= T

2+V

2"

1

2mv

P1

2+ Mg R

1# R

2( ) =1

2mv

P2

2+

1

2Mv

B2

2

Kinematics

vP2

= !R2e

R+ R

2!!e! " v

P2

2= !R

2

2+ R

2!!2( )

2

vB2

= !R2

Solving

FB

O

P

eR

e!

R

B

2

1

2mv

1

2+ Mg R

1! R

2( ) =1

2m !R

2

2+ R

2!"2( )

2#$%

&'(+

1

2M !R

2

2=

1

2m R

2!"2( )

2

+1

2m + M( ) !R2

2 )

!R2= ±

mv1

2+ 2Mg R

1! R

2( ) ! m R2!"2( )

2

m + M= ±

v1

2+ 2 M / m( )g R

1! R

2( ) ! R2!"2( )

2

1+ M / m

= ±3

2+ 2( ) 8 / 2( ) 9.806( ) 2 !1.5( ) ! 1.5( )

2

2.67( )2

1+ 8 / 2( )= ±2.54 m / sec

! v

2= !R

2e

R+ R

2!"2e" = ±2.54e

R+ 4.0e"( ) m / sec

3



PROBLEM NO. 2

Given: A mechanism is made up of links OA and

AB. OA (having negligible mass and

length 0.6L) is pinned to ground at O and

pinned to AB at A. Link AB (having a

mass of m with uniform mass distribution

and length L) is free to slide within a

smooth slot at end B with this slot being

perpendicular to line OD. A force F acts

in the direction shown at B. The system is released from rest when " = 90°.

The mechanism moves within a HORIZONTAL plane.

Find: Determine the angular acceleration of link AB on

release. Use the following parameters: m = 40 kg, L = 2

meters and F = 200 newtons.

FBD of AB – shown to right

Newton-Euler

1( ) Fx! = "N = ma

Gx

2( ) Fy! = "F

OA+ F = ma

Gy

3( ) MG! = F

OA+ F( )

L

2cos#

$%&

'()" N

L

2sin#

$%&

'()= I

G*

AB=

mL2

12*

AB

(1), (2) and (3): 2F cos! " ma

Gycos! + ma

Gxsin! = mL#

AB/ 6

Kinematics

a A = aA

i = aB +! AB " r A/ B #$AB2 r A/ B = a

Bj + !

ABk( ) " #Lcos%i + Lsin% j( )

= #L!AB

sin%( ) i + aB# L!

ABcos%( ) j & i : a

A= #L!

ABsin%

aG = a A +! AB " rG / A #$ 2rG / A = #L!AB

sin%( )i + !AB

k( ) " L / 2( ) cos%i # sin% j( )= # L / 2( )! AB

sin%&' () i + L / 2( )! ABcos%&' () j

Solve

2F cos! " m L / 2( )# ABcos

2! " m L / 2( )# ABsin

2! = mL#AB

/ 6 $

2F cos! = mL 1 / 6 + 1 / 2( ) cos2! + sin

2!( )%&'

()*#

AB= 2 mL / 3( )# AB

$

#AB

= #AB

k =3

mLF cos!k =

3

40( ) 2( )200( ) 0.8( )k = 6 rad / sec

2( )k

x

y

A

!

G

B FOA

F

N

4



PROBLEM NO. 3

Given: The homogeneous solid cylindrical pulley,

shown below, has mass m and radius r. The

attachment at B undergoes the indicated

harmonic displacement and the cord which

connects the mass of 2m does not slip on the

pulley.

Find: For this problem,

a) derive the differential equation of

motion for the system in terms of the

variable x shown below.

b) determine the frequency " for which the system will be at resonance.

Use m = 2 kg, k = 1000 N/m and r = 0.1 meters.

c) determine the amplitude of response at resonance. Justify your answer.

FBDs – shown to right

Newton-Euler

pulley : MO! = Tr " k r# " x

B( )r = IO!!# = mr2!!# / 2

block : Fx! = "T " kx + 2mg = 2m!!x

Combining:

T = k r! " xB( ) + mr !!! / 2 = "kx + 2mg " 2m!!x #

m r !!! / 2 + 2!!x( ) + k r! + x( ) = 2mg + kxB

Kinematics: ! = x / r and !!! = !!x / r

EOM: 5m / 2( ) !!x + 2kx = 2mg + kbsin!t

Natural frequency:

!n=

2k

5m / 2=

2( ) 1000( )5( ) 2( ) / 2

= 20 rad / sec

Response (w/o weight):

xP

t( ) = Asin!t " #! 25m / 2( ) + 2k$

%&'

A = kb "

A =kb

2k #! 25m / 2( )

=b / 2

1# ! /!n( )

2= ( when! =!

n

Ox

Oy

T

k(r# – xB)

T

2mg kx

5



PROBLEM NO. 4

A homogeneous drum of mass m and outer radius R is supported by a

flexible cable, as shown to the right. The drum is known to not slip on

the cable. A constant force F acts at end A of the cable. The coordinate

x (measured positively upward) describes the position of the center of

mass G of the disk. The angle " (measured positively CW) describes

the rotation of the disk.

Problem I

If R = 0.5 ft and !!! = 2 rad / sec , what is !!x ?

!!x = +R!!! = 0.5( ) 2( ) = +1 ft / sec

Problem II

If F = 50 lb and !x = -2ft, what is the work done by F?

U

F= !F 2! x( ) = ! 50( ) 2( ) 2( ) = !200 ft ! lb

Problem III

If mg = 200 lb and !x = -2ft, what is the net change in potential energy, V2 – V1, for the

disk?

V

2!V

1= mg ! x = 200( ) !2( ) = !400 ft ! lb

Problem IV

If the disk starts from rest, and if F = 50 lb, R = 0.5 ft and mg = 200 lb, what is the

angular speed of the drum when !x = -2ft?

UF= T

2+V

2!V

1=

1

2IC" 2

+V2!V

1; I

C= I

G+ mR

2= mR

2/ 2 + mR

2= 3mR

2/ 2

" =4

3mR2

UF! V

2!V

1( )#$

%& =

4

3 200 / 32.17( ) 0.5( )2

!200 ! !400( )#$

%& = 13.1 rad / sec

A

x #

no slip

R

F

G

g

m

6



PROBLEM NO. 4 (continued)

Problem V

The radius of gyration about point A of a homogeneous disk of mass m and

outer radius R is:

a) k

A= R

b) k

A= R / 2

c) k

A= R / 2

d) k

A= 2 / 3 R

e) k

A= 3 / 2 R

Problem VI

A disk is placed on a rough, stationary horizontal surface with its

center moving to the left with a speed of vO and with # = vO/R.

Circle the answer below that most accurately describes the work done

on the disk by friction as the disk moves to the left.

a) Uf > 0

b) Uf = 0

c) Uf < 0

vO

O

rough

R

"

G

R

A

7




Problem VII

A thin bar of mass m slides along rough vertical and horizontal surfaces at A and B, respectively.

The FBD of the bar is shown below right. Circle ALL correct expressions for the kinetic energy

of the bar:

a) T = mv

A

2/ 2 + I

A!!

2/ 2

b) T = mv

G

2/ 2 + I

G!!

2/ 2

c) T = mv

B

2/ 2 + I

B!!

2/ 2

d) T = I

C!!

2/ 2

Problem VIII

A disk moves to the right without slipping and

with its center having a constant speed of vO.

Circle the figure below that most accurately

describes the VELOCITY of point A on the disk.

C

#

G

A

B

vA

vB

vO O

no slip

A

R

r

vA

O

A

vA O

A

vA

O

A

vA

O

A

8




Problem IX

An inhomogeneous disk (of mass m, and with its center of mass

at G and geometric center at O) moves to the right without

slipping on a rough horizontal surface. Circle the answer below

that most accurately describes the size of the normal contact

force N when G is directly below O:

a) N > mg

b) N = mg

c) N < mg

Problem X

A homogeneous disk of mass m and outer radius r travels along a circular surface having a radius

of R. The maximum speed vO for which the disk does not lose contact with the surface at the

position shown is:

a) v

O= 0

b) v

O= gR

c) v

O= g R + r( )

d) v

O= gRcos!

e) v

O= gRsin!

f) v

O= g R + r( )cos!

g) v

O= g R + r( )sin!

N

O G

f

mg vO

vO

O

no slip

G

vO

O

R

r

g

#

December 18, 2009

me 274 – spring 2008 solution final examination ... · pdf file4 me 274 – spring...

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