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

The correct option is D $8\omega$

By conservation of angular momentum $I_1{\omega }_1=I_2{\omega }_2$
${\omega }_2=\frac{I_1{\omega }_1}{I_2}=\frac{\left(m{r}^2\right)\left(2\omega \right)}{m{\left(\frac{r}{2}\right)}^2}$

[changed radius $=\frac{r}{2}$]

$\therefore {\omega }_2=8\omega$