Assistant Professor: Zhou Yufeng
N3.2-01-25, 6790-4482, yfzhou@ntu.edu.sg
http://www3.ntu.edu.sg/home/yfzhou/courses.html
1. The spring of constant k is unstretched when the slider of
mass m passes position B. If the slider is released from
rest in position A, determine its speed as it passes point B
and C. What is the normal force exerted by the guide on
the slider at position C? Neglect friction between the mass
and the circular guide, which lies in a vertical plane.
The system is a conservative system, for position A, B and C we have
E1  E2  E3  T1  V1  T2  V2  T3  V3
1

2
T

V

0

mgR

k
(
2
R

R
)
1 1
2

1 2
T

V

mv2  0

2
2
2

1

T3  V3  mv32  mgR

2
At C:
v32
N  mg  m
R
kR2
v2  2 gR  0.1716
m

kR2
v3  4 gR  0.1716
m
 N  5mg  0.1716kR
2. The pendulum shown is released from rest at A and
swings through 90 before the cord touches the fixed
peg B. Determine the smallest value of a for which the
pendulum bob will describe a circle about the peg.
Assume at position C the tension in the cord is 0, then
vC2
mg  m
r
 vC2  g ( L  a)
Using EA = EC gives
TA  VA  TC  VC
 0  mg (2a  L) 
 g ( 2a  L ) 
 a  0.6 L
1 2
mvC  0
2
1
g ( L  a)
2
3. The block B has a mass of 50 kg and rests on the
surface of the cart having a mass of 75 kg. If the
spring, which is attached to the cart but not the block,
is compressed 0.2 m, the system is released from rest,
determine (a) the velocity of the block after the spring
becomes undeformed, (b) the velocity of the block
relative to the cart after the spring becomes
undeformed. Neglect the mass of the cart’s wheels and
the spring in the calculation. Also neglect friction. Take
k = 300 N/m.
Consider Cart C and Block B and the spring as the system
The system is energy conservative (why?): E1 = E2
1
E1  T1  V1  0  k2
2

1
1
E2  T2  V2  mB vB2  mC vC2  0
2
2
1
1
1
(50)vB2  (75)vC2  0  0  (300)0.2 2
2
2
2
(1)
The linear momentum of the system is conservative (why?):
Assuming vB  v B i , vC   v C i we have
mB vB  mC vC  0  (50)vB  (75)vC  0
Solving (1) and (2) yields
vB  0.379 (m/s),
vC

vB  0.379i (m / s),

 0.253 (m/s)

vC  0.253(m / s)

vB / C  0.632i(m / s)


L1  L2
4. Particle P of mass 4 kg is placed on the smooth
inclined surface of a triangular block B of mass
20 Kg as shown. If the system is released from
rest, determine velocities of P and B when P
reaches the bottom. Neglect the size of particle P
and the friction at all surfaces.

vB  vBiˆ

vP  vBiˆ  vP / B   30
Conservation of Linear momentum in iˆ
 20vB  4(vB  vP / B cos 30)  0
vP2  vB2  vP2 / B  2vP / B vB cos 30
(1)
6
vB , vP2 / B  48vB2
cos 30
1
1
2
2
Principle of Work and Energy: 4 g (2)  (20)vB  (4)vP
2
2
We have vP / B cos 30  6vB , vP / B 
Here
vP2  vB2  vP2 / B  2vP / B vB cos 30  vB2  48vB2  12vB2  37vB2
Solve vB  0.967(m / s), vP / B  6.697(m / s),

vP  vBiˆ  vP / B   30  0.967iˆ  6.697  30(m / s)
(2)
8g  (10  74)vB2