Question

A small block on a frictionless horizontal surface has a mass of 2.50×10−2 Kg. It is attached to a massless cord passing through a hole in the surface. The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 1.75 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. You may treat the block as a particle.
A) is angular momentum conserved?
B) What is the new angular speed?
C) Find the change in kinetic energy of the block.
D) how much work was done in pulling the cord.

For C i have:
KEf=(1/2)(2.5*10^-2)(.150*7)^2
=.01378125

KEi=(1/2)(2.5*10^-2)(.300*1.75)^2
=.0034453125

Answer 1

That's correct. Try to think about why angular momentum is conserved and why kinetic energy is not conserved.

You can also consider a similar problem where the cord is not pulled but instead it is wrapping itself around a stick. What happens to angular momentum and kinetic energy in this case?

Answer 2

how do you find B and D then?

Answer 3

B: form conseravtion of momentum, which yields the angular velocities of 7 rad/sec that you used.

D follows from conservation of energy. The work done equals the increase in kinetic energy.

Answer 4

Typo: momentum --> angular momentum