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. $2$
• C. Zero
• D. $1$

Answer 1

The correct option is A $-1$

Given,
$\overrightarrow{F}=\alpha \hat{i}+3\hat{j}+6\hat{k}$..... $\left(1\right)$
$\overrightarrow{r}=2\hat{i}-6\hat{j}-12\hat{k}$ ....$\left(2\right)$
We know for angular momentum to remain conserved, torque of force has to be zero.
$\Rightarrow \overrightarrow{\tau }=0$
$\Rightarrow \overrightarrow{r}\times \overrightarrow{F}=0$ $[\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}]$
$\Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -6& -12\\ \alpha & 3& 6\end{array}\right|=0$
$\Rightarrow \hat{i}(-36+36)-\hat{j}(12+12\alpha )+\hat{k}(6+6\alpha )=0$
$\Rightarrow 12+12\alpha =0$ or $6+6\alpha =0$
$\Rightarrow \alpha =-1$
The value of $\alpha$ is $-1$ for which angular momentum about origin is conserved.