Question
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 the plane as shown.
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 the plane as shown.
The tension in the string is increased gradually and finally $$m$$ moves in a circle of radius $$\frac{{{R_0}}}{2}.$$ The final value of the kinetic energy is
A.
$$mv_0^2$$
B.
$$\frac{1}{4}mv_0^2$$
C.
$$2mv_0^2$$
D.
$$\frac{1}{2}mv_0^2$$
Answer :
$$2mv_0^2$$
Solution :
Conserving angular momentum
$$\eqalign{ & {L_i} = {L_f} \cr & \Rightarrow m{v_0}{R_0} = mv'\left( {\frac{{{R_0}}}{2}} \right) \cr & \Rightarrow v' = 2{v_0} \cr} $$
So, final kinetic energy of the particle is
$$\eqalign{ & {K_f} = \frac{1}{2}m{{v'}^2} = \frac{1}{2}m{\left( {2{v_0}} \right)^2} \cr & = 4\frac{1}{2}mv_0^2 = 2mv_0^2 \cr} $$
Conserving angular momentum
$$\eqalign{ & {L_i} = {L_f} \cr & \Rightarrow m{v_0}{R_0} = mv'\left( {\frac{{{R_0}}}{2}} \right) \cr & \Rightarrow v' = 2{v_0} \cr} $$
So, final kinetic energy of the particle is
$$\eqalign{ & {K_f} = \frac{1}{2}m{{v'}^2} = \frac{1}{2}m{\left( {2{v_0}} \right)^2} \cr & = 4\frac{1}{2}mv_0^2 = 2mv_0^2 \cr} $$