System of Particles and Rotational Motion - Result Question 23
23. 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 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 [2015]
(a) $\frac{1}{4} mv_0^{2}$
(b) $2 mv_0^{2}$
(c) $\frac{1}{2} mv_0^{2}$
(d) $mv_0^{2}$
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Answer:
Correct Answer: 23. (b)
Solution:
- (b) Applying angular momentum conservation
$mv_0 R_0=(m)(v)(\frac{R_0}{2})$
$\therefore \quad v^{\prime}=2 v_0$
Therefore, final $KE=\frac{1}{2} m(2 v_0)^{2}=2 mv_0^{2}$
