A small block on a frictionless horizontal surface has a mass of 2.15×10 2 kg .
ID: 1464115 • Letter: A
Question
A small block on a frictionless horizontal surface has a mass of 2.15×102 kg . It is attached to a massless cord passing through a hole in the surface. (See the figure below (Figure 1) .) The block is originally revolving at a distance of 0.265 m from the hole with an angular speed of 1.95 rad/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.165 m . You may treat the block as a particle.
A)Is angular momentum conserved
B)What is the new angular speed
C)Find te change in kinetic energy of the block
D)How much work was done in pulling the cord?
Explanation / Answer
A)
yes the angular momentum is conserved
B)
m = mass of small block = 0.0215 kg
r1 = initial distance from hole = 0.265 m
r2 = final distance from hole = 0.165 m
W1 = initial angular velocity = 1.95 rad/s
W2 = final angular velocity
Using conservation of angular momentum
m r12 W1 = m r22 W2 mr2 = moment of inertia
(0.265)2 (1.95) = (0.165)2 W2
W2 = 2.24 rad/s
c)
initial rotational KE = (0.5) m r12 W21
final rotational KE = (0.5) m r22 W22
change in KE = (0.5) m r12 W21 - (0.5) m r22 W22 = (0.5) m (r12 W21 - r22 W22 )
change in KE = (0.5) (0.0215) ((0.265)2 (1.95)2 - (0.165)2 (2.24)2 ) = 0.00140
d)
work done = change in kinetic energy = 0.00140