Question

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a radius of 3.87m and a rotational inertia of 429kg.m^2 about the axis of rotation. A 78.4kg student walks slowly from the rim of the platform towards the center. The angular speed of the system is 2.95rad/s when the student starts at the rim.

1. What is the angular speed when the student is 0.833m from the center? (this is not at half radius.)
2. How much kinetic energy has been lost in the system?

Answer 1

The moment of inertia of the disk w.r.t rotational axis is

The moment of inertia of the student w.r.t. rotational axis when he is at the rim of the disk is

$I_{s}=m_sr_d^2=78.4\times 3.87^2=1174.19\ kg.m^2$

The total moment of inertia of the system when the student is at the rim is

The initial angular momentum is

Li = lw= 1603.19 x 2.95 = 4729.41 kg.ma/s

And the initial kinetic energy of the system is

$K_i=\frac{1}{2}I\omega^2=\frac{1}{2}\times 1603.19\times2.95^2=6976\ J$

Now, when the student is at 0.883 m from the center, the moment of inertia of the student w.r.t. the axis of rotation is

$I_{s}'=m_sr^2=78.4\times 0.883^2=61.13\ kg.m^2$

So, the total moment of inertia of the system is

Now, the angular momentum of the system is

$L_f=I'\omega_f=490.13\omega_f$

Conservation of angular momentum gives us

$L_f=L_i$

$\Rightarrow 490.13\omega_f=4729.41$

$\Rightarrow \omega_f=\frac{4729.41}{490.13}=9.65\ rad/s$

So, the final angular velocity is 9.65 rad/s.

Now, the final kinetic energy of the system is

$K_f=\frac{1}{2}I'\omega_f^2=0.5\times 490.13\times 9.65^2= 22820\ J$

It is notable that the kinetic energy of the system has been increased.

And the increase in the kinetic energy is

$\Delta K=K_f-K_i=22820-6976=15844\ J$

So, the total energy put in the system is 15844 J.