Question

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

• A. $\text{Zero}$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is C $-1$

Since position vector of the point with respect to origin will be:
$\overrightarrow{r}=2\hat{i}-6\hat{j}-12\hat{k}$
Equation of torque for a force $\overrightarrow{F}$ acting at a point whose position vector is $\overrightarrow{r}$ about a reference point
$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -6& -12\\ \alpha & 3& 6\end{array}\right|$
$\overrightarrow{\tau }=\hat{i}\left[\right(-6\times 6)-(-12\times 3\left)\right]-\hat{j}\left[\right(2\times 6)-(-12\times \alpha \left)\right]+\hat{k}\left[\right(2\times 3)-(-6\times \alpha \left)\right]$
$\Rightarrow \overrightarrow{\tau }=0\hat{i}-(12+12\alpha )\hat{j}+(6+6\alpha )\hat{k}$
$\Rightarrow \overrightarrow{\tau }=0\hat{i}-(12+12\alpha )\hat{j}+(6+6\alpha )\hat{k}$

For angular momentum to be conserved about origin,
$\therefore {\tau }_{\text{ext}}=0$ about origin
$\Rightarrow 0\hat{i}-(12+12\alpha )\hat{j}+(6+6\alpha )\hat{k}=0$

Equating the coefficients of $\hat{i},\hat{j},\hat{k}$ respectively to $\text{Zero}$, we get:
$\Rightarrow 12+12\alpha =0$ and $6+6\alpha =0$
$\therefore \alpha =\frac{-12}{12}=\frac{-6}{6}=-1$