Question

A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm). The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm) of the platform when the person reaches its centre _____?

Answer 1

Hint: Recall conservation of angular momentum and the concept of moment of inertia. If you know the moment of inertia about the axis and about the rim and if you know the equation of angular momentum you can easily solve this problem. Moment of inertia is the measure of rotational inertia of a body.



Formula used:
From conservation of angular momentum,
$I\omega $ =constant.
Where,
I=moment of inertia
$\omega$=angular frequency
and we use another equation
$\omega =2\pi f$
where f=frequency.


Complete answer:
We have the equation:
Moment of inertia, $I=\sum\limits_{i}{{{m}_{i}}}r_{i}^{2}$
Where ${m_i}$ is the mass of ‘i’th particle and ${r_i}$ is its perpendicular distance from the given line.

If the system is considered to be the collection of discrete particles, this definition may be directly used to calculate the moment of inertia. Moment of inertia depends on the mass of the body, axis of rotation of the body and the size and shape of the body.

Here in this question, first the person is standing on the rim of the circular platform and in the second case the person is standing at the centre of the circular platform. The moment of inertia about the rim and centre of the circular platform is different but angular momentum at these two positions is conserved.

When conservation of angular momentum is applied, we have:
${{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}$ -(1)
Where $I_{1}$ and $\omega_{1}$ is the moment of inertia and angular velocity of the person when person stands on the rim of the circular platform and $I_{2}$ and $\omega_{2}$ is the moment of inertia and angular velocity of the person when the person stands at the centre of circular platform.
We have,
${{I}_{1}}=\dfrac{M{{R}^{2}}}{2}+m{{R}^{2}}$
where M is the mass of a circular platform and m is the mass of the person.
${{I}_{2}}=\dfrac{M{{R}^{2}}}{2}$
On substituting the values of moment of inertia in the equation (1)
We get $\frac{{{\omega }_{2}}}{{{\omega }_{1}}}=1+\frac{2\times 80}{200}=1.8 $
We have \[\omega =2\pi f\]
$\therefore \frac{{{\omega }_{2}}}{{{\omega }_{1}}}=\frac{{{f}_{2}}}{{{f}_{1}}}=1.8$
$\therefore {{f}_{2}}=5\times 1.8=9rpm$
Therefore, the answer is: 9 revolutions per minute(rpm)





Note: You have to remember the moment of inertia about the axis and about the rim is different.