Question

A small steel sphere of mass $m$ is tied to a string of length $r$ and is whirled in a horizontal circle with a uniform angular velocity $ 2\,\omega $ . The string is suddenly pulled, so that radius of the circle is halved. The new angular velocity will be:

• A. $ 2\omega $
• B. $ 4\omega $
• C. $ 6\omega $
• D. $ 8\omega $

Answer 1

Answer: (D)

Given : Mass of sphere $=m$
Initial radius of path $r_{1}=r$
Final radius of path $r_{2}=\frac{r}{2}$
The initial M.I. of a particle of mass $m$ rotating in a horizontal circle is given by
$=m r_{1}^{2}=m r^{2}$
Similarly, the final M.I. of a particle rotating in a circle is
$=m r_{2}^{2}=m\left(\frac{r}{2}\right)^{2} $
$=\frac{m r^{2}}{4}$
Applying law of conservation of momentum
$ m r^{2} \times 2\, \omega=\frac{m r^{2}}{4} \times \omega_{2}$
$\therefore \omega_{2}=4 \times 2\, \omega=8 \,\omega$