Question

the torque $\overrightarrow{r}$ 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 that the point. From the it follows that

• A. $\frac{dL}{dt}$ 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. L does not change with time

Answer 1

The correct option is C The magnitude of $\overrightarrow{L}$ does not change with time

Given $\tau =\overrightarrow{A}\times \overrightarrow{L}$

$\overrightarrow{L}$ is angular momentum of the body.

we know that $\tau =\frac{d\overrightarrow{L}}{dt}$

substituting above value in place of $\tau$,

$\tau =\overrightarrow{A}\times \overrightarrow{L}=$ $\frac{d\overrightarrow{L}}{dt}$

and we know cross product of any two vector is perpendicular to both vector

which means $\frac{d\overrightarrow{L}}{dt}$ is perpendicular to $\overrightarrow{L}$ and $\overrightarrow{A}$.

cross product can be expanded as

$\frac{d\overrightarrow{L}}{dt}=Al\sin \theta$ where $\theta$ is angle between $\overrightarrow{L}$ and $\overrightarrow{A}$.

component of $\overrightarrow{L}$ in direction of $\overrightarrow{A}$ is not perpendicular to torque and it will change with time.

Torque acting is perpendicular to L so magnitude of L will not change with time.

Hence option (C) is correct.