Question

A force $\overrightarrow{F}=\alpha \hat{i}+3\hat{j}+6\hat{k}$ is acting at a point $\overrightarrow{r}=2\hat{i}-6\hat{j}-12\hat{k}$. The value of $\alpha$ for which angular momentum about origin is conserved is:

• A. 1
• B. -1
• C. 2
• D. zero

Answer 1

The correct option is B -1

Rate of change of angular momentum is torque applied.
$\overrightarrow{\tau }=\frac{\overrightarrow{dL}}{dt}=0$ since $\overrightarrow{L}$ is constant.

$\therefore \overrightarrow{r}\times \overrightarrow{F}=(2\hat{i}-6\hat{j}-12\hat{k})\times (\alpha \hat{i}+3\hat{j}+6\hat{k})=0$

$\Rightarrow \hat{i}(-36+36)+\hat{j}(-12\alpha -12)+\hat{k}(6+6\alpha )=0$

$\Rightarrow -12\alpha -12=0$

$\Rightarrow \alpha =-1$