A small block on a frictionless horizontal surface has a mass of 0.0300 kg . It
ID: 1418928 • Letter: A
Question
A small block on a frictionless horizontal surface has a mass of 0.0300 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 0.350 m from the hole with an angular speed of 1.50 rad/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.175 m . You may treat the block as a particle.
What is the new angular speed?
Find the change in kinetic energy of the block
Explanation / Answer
Using angular momentum conservation:
L1 = L2
I1*w1 = I2*w2
I = m*r^2
m*r1^2*w1 = m*r2^2*w2
w2 = (r1/r2)^2*w1
w2 = (0.350/0.175)^2*1.5
w2 = 6 rad/sec
B.
KE1 = 0.5*I1*w1^2 = 0.5*m*r1^2*(v1/r1)^2
KE1 = 0.5*m*v1^2
v1 = r1*w1 = 0.350*1.5 = 0.525 m/sec
v2 = r2*w2 = 0.175*6 = 1.05 m/sec
KE1 = 0.5*0.03*0.525^2 = 4.13*10^-3 J
KE2 = 0.5*0.03*1.05^2 = 16.54*10^-3 J
dKE = KE2 - KE1 = (16.54 - 4.13)*10^-3
dKE = 12.41*10^-3 J = 0.01241 J