A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius R02. [figure] The final value of the kinetic energy is
- 2mv02
- 12mv02
- mv02
- 14mv02
Answer
(1) 2mv02
Solution
The tension acts along the string, towards the hole, so its torque about the hole is zero.
Angular momentum of the mass about the hole is conserved:
mv0R0 = mv ( R02 )
v = 2v0
Kf = 12m(2v0)2 = 2mv02
Answer: (1)