Question

A mass is whirled in a circular path with a constant angular velocity and its angular momentum is $L$. If the length of string is now halved keeping the angular velocity the same, the new angular momentum is:

• A. $\frac{L}{4}$
• B. $\frac{L}{2}$
• C. $L$
• D. $2L$

Answer 1

The correct option is A $\frac{L}{4}$

Angular momentum, $L=mvr$

Also $v=r\omega ⟹L=m{r}^2\omega$

Now, $r_2=\frac{r}{2}$ (putting it in the earliar equation to get new momentum)

Thus $L_2=m\frac{r^2}{4}\omega$

$⟹L_2=\frac{L}{4}$