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 $\dfrac{M}{m}=4$)

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 =$\dfrac{m{{r}^{2}}}{2}$, 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
${{L}_{1}}=(\dfrac{M{{R}^{2}}}{2}+m{{(0)}^{2}}){{\omega }_{2}}$ (since particle is at centre its distance from axis is 0)
${{L}_{1}}=\dfrac{M{{R}^{2}}{{\omega }_{2}}}{2}$

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

Angular momentum in final condition,

${{L}_{2}}=(\dfrac{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 ${{L}_{2}}$
$\dfrac{M{{R}^{2}}{{\omega }_{2}}}{2}=(\dfrac{M{{R}^{2}}}{2}+m{{R}^{2}})\omega $
$\omega =\dfrac{\dfrac{M{{R}^{2}}{{\omega }_{2}}}{2}}{(\dfrac{M{{R}^{2}}}{2}+m{{R}^{2}})}$
Dividing numerator and denominator in RHS by m and putting $\dfrac{M}{m}=4$.
$\omega =\dfrac{2{{R}^{2}}{{\omega }_{2}}}{2{{R}^{2}}+{{R}^{2}}}$
$\omega =\dfrac{2{{\omega }_{2}}}{3}$

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

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