NEET2015PhysicsRotational MotionActual
A mass m moves in a circle on a smooth horizontal plane with velocity v 0 at a radius R 0 . The mass is attached to a string which passes through a smooth hole in plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius R 0 2 . The final value of the kinetic energy is:
Options
- Am v 0 2
- B1 4 m v 0 2
- C2 m v 0 2
- D1 2 m v 0 2
Correct answer
C. 2 m v 0 2
Step-by-step solution
From conservation of angular momentum. ( aligned & m v₀ R₀=m v^ ( R₀ 2 ) & v^ =2 v₀ aligned ) Hence, final (K E= 1 2 m v^ 2 = 1 2 m (2 v₀ )^2 ) Final (KE=2 m v₀^2 )