Question

A rotating table completes one rotation in $10s$, and its moment of inertia is $100kg - {m^2}$. A person of $50kg$ mass stands at the centre of the rotating table. If the person moves 2m from the centre, the angular velocity of the rotating table in ($rad/s$) will be:
A. $\dfrac{{2\pi }}{{30}}$
B. $\dfrac{{20\pi }}{{30}}$
C. $\dfrac{{2\pi }}{3}$
D. $2\pi$

Answer 1

Hint: Final moment of inertia of the man and the table is equal to the sum of individual moments of inertia of the table and the man considering the man's distance from the centre of rotation. Once you find the new moment of inertia use conservation of angular momentum to find the final angular velocity.

Complete answer:
The rotating table takes ${t_1} = 10s$ to complete one rotation. So its angular velocity, ${\omega _1} = \dfrac{{2\pi }}{{{t_1}}} = \dfrac{{2\pi }}{{10}}rad/s$.
Let the given moment of inertia of the table be $I = 100kg - {m^2}$ and the mass of the person initially at the centre of table be $M = 50kg$. Now the person moves 2m from the centre and let this distance be $r = 2m$. Initially the person was standing at the centre of the rotation and thus it was not contributing in the moment of inertia. But as soon as he/she moves a distance away from the centre, his/her mass will start contributing in the moment of inertia. Let the new moment of inertia be ${I_f}$ and it is as follows:
${I_f} = I + M{r^2}$. Since the centre for rotation remains the same, so is the axis of rotation. Therefore the moment of inertia of the table does not change. New moment of inertia is given as follows:
${I_f} = 100 + 50 \times {2^2} = 300kg - {m^2}$
Since there is no external force acting on this complete system, the angular momentum will be conserved and remains constant before and after movement of the person.
Initial angular momentum, $= I{\omega_1}$ and if we consider the final angular velocity as ${\omega _f}$, then
Final angular momentum, $= {I_f}{\omega _f}$ and using the above conservation principle, we get,
$\Rightarrow {I_f}{\omega _f} = I{\omega_1}$
$\Rightarrow \therefore {\omega _f} = \dfrac{{I\omega_1}}{{{I_f}}} = \dfrac{{100}}{{300}} \times \dfrac{{2\pi }}{{10}} = \dfrac{{2\pi }}{{30}}rad/s$
Comparing the above solution with the choices in the question, we say that option B is correct.

Note: Conservation of angular momentum can be considered analogous to conservation of linear momentum, where the moment of inertia is analogous to mass and angular velocity is analogous to linear velocity.