System of Particles and Rotational Motion - Result Question 27
27. 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 :
[2015 RS]
(a) 2
(b) zero
(c) 1
(d) -1
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Answer:
Correct Answer: 27. (d)
Solution:
- (d) From Newton’s second law for rotational motion,
$\vec{\tau}=\frac{d \overrightarrow{{}L}}{dt}$, if $\overrightarrow{{}L}=$ constant then $\vec{\tau}=0$
So, $\vec{\tau}=\overrightarrow{{}r} \times \overrightarrow{{}F}=0$
$(2 \hat{i}-6 \hat{j}-12 \hat{k}) \times(\alpha \hat{i}+3 \hat{j}+6 \hat{k})=0$
Solving we get $\alpha=-1$
