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
ME 274 – Spring 2008 SOLUTION
Final Examination - Wednesday
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
ME 274 – Spring 2008 SOLUTION
Final Examination - Wednesday
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
ME 274 – Spring 2008 SOLUTION
Final Examination - Wednesday
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
ME 274 – Spring 2008 SOLUTION
Final Examination - Wednesday
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
ME 274 – Spring 2008 SOLUTION
Final Examination - Wednesday
PROBLEM NO. 4 (continued)
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
ME 274 – Spring 2008 SOLUTION
Final Examination - Wednesday
PROBLEM NO. 4 (continued)
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
December 18, 2009
December 18, 2009
December 18, 2009
December 18, 2009