Question

A uniform horizontal circular disc of mass M and radius R can freely rotate about a vertical axle passing through its centre. A particle of mass m is placed near the centre of disc in a smooth groove made along the radius of the disc as shown. An initial angular velocity ${\omega }_2$ is imparted to the disc. Find the angular velocity of the disc, when the particle reaches the other end of the groove. (Take ${\displaystyle \frac{M}{m}}=4$)

Answer 1

Hint: Since, no external torque acts on the system there will be no change in angular momentum. Therefore, we can apply conservation of angular momentum.

Formula used: Initial angular momentum $(L_1)$= Final angular momentum$(L_2)$

$I_1{\omega }_1=I_2{\omega }_2$, where,$I_1$and $I_2$ are moment of inertia of system in initial and final situation and

${\omega }_1$and ${\omega }_2$ are angular velocities of the system in initial and final situations.

Moment of inertia for disc =${\displaystyle \frac{m{r}^2}{2}}$, where, m is its mass and r is its radius.
Moment of inertia for a particle =$m{r}^2$, where m is its mass and r is its distance from the axis.

Complete step-by-step answer:
Angular momentum in initial condition,

$(L_1)$= angular momentum of disc + particle
$L_1=({\displaystyle \frac{M{R}^2}{2}}+m{(0)}^2){\omega }_2$ (since particle is at centre its distance from axis is 0)
$L_1={\displaystyle \frac{M{R}^2{\omega }_2}{2}}$

Let angular velocity of disc in final condition be $\omega$.

Angular momentum in final condition,

$L_2=({\displaystyle \frac{M{R}^2}{2}}+m{R}^2)\omega$(since particle reached end of groove its distance from centre will be R)

Equating $L_1$ and $L_2$
${\displaystyle \frac{M{R}^2{\omega }_2}{2}}=({\displaystyle \frac{M{R}^2}{2}}+m{R}^2)\omega$
$\omega ={\displaystyle \frac{{\displaystyle \frac{M{R}^2{\omega }_2}{2}}}{({\displaystyle \frac{M{R}^2}{2}}+m{R}^2)}}$
Dividing numerator and denominator in RHS by m and putting ${\displaystyle \frac{M}{m}}=4$.
$\omega ={\displaystyle \frac{2R^2{\omega }_2}{2R^2+R^2}}$
$\omega ={\displaystyle \frac{2{\omega }_2}{3}}$

Therefore, angular velocity when a particle reaches the end of the groove is ${\displaystyle \frac{2{\omega }_2}{3}}$.

Note: Since, disc and ring are very similar students often consider them as one and instead of using moment of inertia of disc as ${\displaystyle \frac{m{r}^2}{2}}$ they take it as $m{r}^2$which is the moment of inertia of a ring.