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 in the figure. The tension in the string is increased gradually and finally m moves in a circle of radius $$R0$$ $$/2$$. The final value of the kinetic energy is [CBSE AIPMT $$2015$$]

Question

Infinity Learn · Accepted Answer

According to law of conservation of angular momentum,$$Li=Lf$$⇒$$mv0R0=mv$$′$$R02$$⇒v−′$$=2v0So$$, final kinetic energy of the particle,$$Kf=12mv2=12m2v02=4$$×$$12mv02=2mv02$$

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