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 $\vec{F}=\alpha \hat{i}+3 \hat{j}+6 \hat{k}$
$\therefore \vec{\tau}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -12 \\ \alpha & 3 & 6\end{vmatrix}$
$=\hat{i}(-36+36)-\hat{j}(12+12 \alpha)+\hat{k}(6+6 \alpha)$
$=-\hat{j}(12+12 \alpha)+\hat{k}(6+6 \alpha)$
But $\vec{\tau}=0$
$\therefore 12+12 \alpha=0 $ or $ \alpha=-1 $
and $6+6 \alpha=0 $ or$ \alpha=-1$$