Question

Angular momentum of body about a point $A$ is given as $\overrightarrow{L}=(4t^2+6)\hat{k}$. Find the net torque acting on body about point $A$ at $t=3\text{sec}$.

• A. $8\hat{k}\text{Nm}$
• B. $4\hat{k}\text{Nm}$
• C. $24\hat{k}\text{Nm}$
• D. $20\hat{k}\text{Nm}$

Answer 1

The correct option is C $24\hat{k}\text{Nm}$

As we know,
${\overrightarrow{\tau }}_{\text{net}}=\frac{d{\overrightarrow{L}}_{\text{sys}}}{dt}$
Given:
$\overrightarrow{L}=(4t^2+6)\hat{k}$
$\frac{d\overrightarrow{L}}{dt}=8t\hat{k}$
$\therefore$ At $t=3\text{sec}$
$\frac{d\overrightarrow{L}}{dt}=(8\times 3)\hat{k}=24\hat{k}\text{Nm}$
$\therefore$ Option (c) is correct.