Question

A sphere is released from the top of a smooth inclined plane . When it moves downwards, its angular momentum will be

• A. conserved about any point
• B. conserved about the point of contact only
• C. conserved about the centre of the sphere only
• D. conserved about any point on a fixed line parallel to the inclined plane and passing through the centre of the sphere

Answer 1

The correct option is D conserved about any point on a fixed line parallel to the inclined plane and passing through the centre of the sphere


$\overrightarrow{\tau }=\frac{d\overrightarrow{L}}{dt}$
If ${\overrightarrow{\tau }}_{\text{ext}}=0,\frac{d\overrightarrow{L}}{dt}=0$
It means angular momentum is conserved (about the reference point).

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$
where $\overrightarrow{r}=$ position vector of the force $\overrightarrow{F}$ w.r.t the reference point
Since $N=mg\cos \theta$, their torques will cancel out about any point on the line passing through the centre of the sphere $\&$ parallel to the inclined plane. Thus, no torque produced by the pair of $N\&mg\cos \theta$.

Also since the line of action of $mg\sin \alpha$ passes through the centre of sphere and parallel to the incline plane, hence its torque, $\overrightarrow{\tau }=(\overrightarrow{r}\times \overrightarrow{F})=0$.
$\therefore {\overrightarrow{\tau }}_{\text{ext}}=0$ and angular momentum about any point on a line parallel to the inclined plane and passing through the centre of the sphere will be conserved.