Question

# A horizontal disc rotating freely about a vertical axis through its center makes $90$ revolutions per minute. A small piece of wax of mass$m$ falls vertically on the disc and sticks to it at a distance$r$ from the axis. If the number of revolutions per minute reduce to $60$, then the moment of inertia of the disc is?

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Solution

## Step 1. Given dataA horizontal disc rotating freely about a vertical axis through its center makes $90$ revolutions per minute, i.e. ${\omega }_{1}=90rpm$The mass of the wax =$m$ Wax sticks to the disk at a distance of $r$ from the axisAfter wax fell on the disc the new angular velocity is ${\omega }_{2}=60rpm$Let, the moment of inertia of the disc is $I$.Step 2. Formula usedIn this case, the total angular momentum of the system must be conserved.$L=I\omega$is conserved.Where $L$ is the angular momentum, $I$ moment of inertia, $\omega$is angular velocityStep 3. Find the moment of inertiaThe initial angular momentum is ${L}_{1}={I}_{1}{\omega }_{1}=90{I}_{1}$The wax is attached to the disc at a distance of $r$ from axis, so the new moment of inertia of the disc with the wax piece is $I+m{r}^{2}$.The final angular momentum ${L}_{2}=$${\omega }_{2}\left(I+m{r}^{2}\right)=60\left(I+m{r}^{2}\right)$From Law of conservation of angular momentum we have ${L}_{1}={L}_{2}\phantom{\rule{0ex}{0ex}}⇒I{\omega }_{1}=\left(I+m{r}^{2}\right){\omega }_{2}\phantom{\rule{0ex}{0ex}}⇒90I=60\left(I+m{r}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒30I=60m{r}^{2}\phantom{\rule{0ex}{0ex}}⇒I=2m{r}^{2}\phantom{\rule{0ex}{0ex}}$ Hence, the moment of inertia of the disc is $2m{r}^{2}$.

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