Question

A small block with mass m is attached to a cord passing through a hole in a frictionless, horizontal surface. The block is initially revolving at a distance r from the hole with a speed vr. as shown. (a) By what force F, applied by the hand is the block held rotating? The cord is then pulled additionally from below, shortening the radius of the circle in which the block revolves to At this new distance, (b) what will the final speed v2 of the block be? (c) By what force F2 will the block be held rotating now? (d) How much was the total work IV, done on m? (e) How much was the work Wr done by the person who pulled on the cord?

4. A small block with mass m is attached to a cord passing through a hole in a frictionless, horizontal surface. The block is initially revolving at a distance r from the hole with a speed vi, as shown. (a) By what force F, applied by the hand is the block held rotating? The cord is then pulled additionally from below, shortening the radius of the circle in which the block revolves to r/2. At this new distance, (b) what will the final speed v2 of the block be? (c) By what force F2 will the block be held rotating now? (d) How much was the total work W. done on m? (e) How much was the work W: done by the person who pulled on the cord?

Answer 1

Given
mass of the block m , moving on a frictionless horizontal surface, in circular path
initially with velocity v1 and radius r1,
a) In order to held the block rotating there should be a force towards the center called centripetal force is given by

   F1 = m*v1^2/r1
b) if the block is pulled down by the cord then

By conservation of angular momentum

   L1 = L2

   m*v1*r1 = m*v2*r2

   v1*r1 = v2*r2
r2 = r1/2
   v1*r1 = v2*r1/2

   v2 = 2*v1

c) the force required is

   F2 = m*v2^2/r2

   F2 = m*4*v1^2/(r1/2)
   F2 = 8 m*v1^2/r1

   F2 = 8*F1
d) Total work done on m is W

By work energy theorem

   W = change in kinetic energy

   W = 0.5*m(v2^2-v1^2)