Question

A turntable has moment of inertia 1 kg m² and rotates with angular speed of 3 rad/s. A small heavy mass (point particle) of mass 5 kg is placed on the turn table at 2 m from the axis of rotation of the turn table. What is the final rotational velocity of the system? Give your result in rad/s to three significant figures.

Answer 1

Angular momentum = I*, where I is moment of inertia and is angular velocity.

So, initial angular momentum = 1*3 = 3 kgm2/s.

Now, moment of inertia of a point mass=mr^2, where m is mass and r is its position relative to the axis.

So, final moment of inertia = 1+5*2*2 = 21 kg m2

Let the final angular velocity be .

So, final angular momentum = 21

Now, according to law of conservation of angular momentum, if there is no net external torque on the system then angular momentum of the system is conserved.

So,21=3

=>=3/21=0.143 rad/s.