Question

The torque $\tau$ on a body about a given point is found to be equal to $\overrightarrow{A}\times \overrightarrow{L}$, where $\overrightarrow{A}$ is a constant vector and $\overrightarrow{L}$ is the angular momentum of the body about the point. From this, it follows that $\tau =\frac{d\overrightarrow{L}}{dt}$ (where $L$ is the angular momentum and $t$ is time).
a) $\overrightarrow{A}$ is perpendicular to $\overrightarrow{L}$ at all instants of time.
b) The component of $\overrightarrow{L}$ in the direction of $\overrightarrow{A}$ does not change with time.
c) The magnitude of $\overrightarrow{L}$ does not change with time.
d) $\overrightarrow{L}$ does not change with time.

• A. a, d are true
• B. b, c, d are true
• C. b, c are true
• D. a,c, d are true

Answer 1

Answer: (C)

b, c are true

Torque $\tau =\frac{d\overrightarrow{L}}{dt}$ where $L$ is angular momentum.
$\tau =\overrightarrow{A}\times \overrightarrow{L}$

Taking dot product on either side with $\overrightarrow{L}$ ,
$\Rightarrow \overrightarrow{L}.\frac{d\overrightarrow{L}}{dt}=0$
$\Rightarrow \overrightarrow{L}.d\overrightarrow{L}=0\Rightarrow d\overrightarrow{L}=0$

Magnitude of $\overrightarrow{L}$ thus remains constant.
Let the component of $\overrightarrow{L}$ in the direction of $\overrightarrow{A}$ be $\overrightarrow{L_{\parallel }}$ and perpendicular to $\overrightarrow{A}$ be $\overrightarrow{L_{\perp }}$.
$\frac{d(\overrightarrow{L_{\parallel }}+\overrightarrow{L_{\perp }})}{dt}=\overrightarrow{A}\times (\overrightarrow{L_{\parallel }}+\overrightarrow{L_{\perp }})$

So, $\frac{d\overrightarrow{L_{\parallel }}}{dt}=\overrightarrow{A}\times \overrightarrow{L_{\parallel }}=0$

Hence $\overrightarrow{L_{\parallel }}$ that is component of $\overrightarrow{L}$ in the direction of $\overrightarrow{A}$ is always constant.