A mass tied to a string is whirled in a horizontal circular path with a constant angular velocity and its…
A mass tied to a string is whirled in a horizontal circular path with a constant angular velocity and its angular momentum is L. if the string is now halved, keeping angular velocity same, then the angular momentum will be
L
$\frac{\mathrm{L}}{4}$
$2 \mathrm{~L}$
$\frac{\mathrm{L}}{2}$
Solution
The correct option is (B).
Concept: Angular momentum is given by $\mathrm{L}=\mathrm{m} \omega \mathrm{r}^2$, where $\mathrm{m}$ is the mass, $\omega$ the angular velocity and $\mathrm{r}$ is the radius of the circular motion (length of the string).
If the new length is half of the original string, then the new angular momentum would be,
$\mathrm{L}_{\mathrm{n}}=\mathrm{m \omega}\left(\frac{\mathrm{r}}{2}\right)^2$
Or
$\mathrm{L}_{\mathrm{n}}=\left(\mathrm{M} \omega \mathrm{r}^2\right)=\frac{1}{4}=\frac{\mathrm{L}}{4}$