Question

The torque $\tau$ on a body about a given point is found to be equal to A x L, where A is a constant vector and L is the angular momentum of the body about that point. From this it follows that

• A. $\frac{dL}{dt}$ is perpendicular to L at all instants of time
• B. the component of L in the direction of A does not change with time
• C. the magnitude of L does not change with time
• D. L does not change with time

Answer 1

Answer: (A), (B), (C)

$(a)\tau =A\times L$
i.e $\frac{dL}{dt}=A\times L$
This relation implies that $\frac{dL}{dt}$is perpendicular to both A and L Therefore, option (a) is correct.
(c) Here, $LL=L^2$
Differentiating w.r.t. time, we get
$L.\frac{dL}{dt}+\frac{dL}{dt}.L=2L\frac{dL}{dt}$
$\Rightarrow 2L.\frac{dL}{dt}=2L\frac{dL}{dt}$
But since, $L-\frac{dL}{dt}$
$\therefore L.\frac{dL}{dt}=0$
Therefore, from Eq. (i) $\frac{dL}{dt}=0$ or magnitude of L i.e. L does not change with time,
(b) So far we are confirm about two points
(1) $\tau or\frac{dL}{dt}\perp L$ and
(2) | L | = L is not changing with time, therefore, it is a case
when direction of L is changing but its magnitude is constant and x is perpendicular to L at all points.
This can be written as
If $L=(acos\theta )\hat{i}+(asin\theta )\hat{j}$
Here, a = positive constant
Then, $\tau =(asin\theta )\hat{i}-(acos\theta )\hat{j}$
So, that $L\tau =0andL\perp \tau$
Now, A is constant vector and it is always perpendicular to x.
Thus, A can be written as A = A $\hat{k}$
we can see that L $\bullet$ A = 0 i.e. L$\perp$ A also.
Thus, we can say that component of L along A is zero or component of L along A is always constant.
Finally we conclude that $\tau$, A and L are always mutually perpendicular.