Physics - Rotation Question with Solution | TestHub

The angular momentum of a particle of mass $m$ moving in a circle of radius $r$ with angular velocity $\omega$ is given by $L=I\omega =m{r}^2\omega$.

Given that the initial angular momentum is $L=m{r}^2\omega$.

When the length of the string is halved, the new radius becomes $r^{\prime }=\frac{r}{2}$.

The angular velocity $\omega$ is kept constant.

The new angular momentum is $L^{\prime }=m(r^{\prime }{)}^2\omega =m{\left(\frac{r}{2}\right)}^2\omega =\frac{m{r}^2\omega }{4}$.

Substituting the initial angular momentum, we get $L^{\prime }=\frac{L}{4}$.

Answer: $\frac{L}{4}$