Question

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

• A. Zero
• B. 1
• C. -1
• D. 2

Answer 1

Answer: (C)

For the conservation of angular momentum about origin, the torque $\vec{\tau }$ acting on the particle will be zero.
By definition, $\vec{\tau }=\vec{r}\times \vec{F}$
Here, $\vec{r}=2\hat{i}-6\hat{j}-12\hat{k}$ and $\vec{F}=\alpha \hat{i}+3\hat{j}+6\hat{k}$
$\therefore \vec{\tau }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -6& -12\\ \alpha & 3& 6\end{array}\right|$
$=\hat{i}(-36+36)-\hat{j}(12+12\alpha )+\hat{k}(6+6\alpha )$
$=-\hat{j}(12+12\alpha )+\hat{k}(6+6\alpha )$
But $\vec{\tau }=0$
$\therefore 12+12\alpha =0$ or $\alpha =-1$
and $6+6\alpha =0$ or$\alpha =-1$$