A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 1.1 m from the axis of rotation, and he rotates with angular speed of 0.64 rad/sec. The moment of inertia of the student plus the stool is 4 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.33 m from the rotation axis. Required:
a. Find the new angular speed of the student.
b. Find the kinetic energy of the student before and after the objects are pulled in.

a) ω₁ = 0.97 rad/sec

b) K₀ = 1.31 J

   Kf  = 1.99 J

Explanation:

a)

Assuming no external torques acting on the student, total angular momentum must be conserved, as follows:

       

The angular momentum of a rigid body rotating respect an axis of rotation can be written as follows:

       

In order to get ωf, since ω₀ = 0.64 rad/sec, we need to find the values of the initial moment of inertia, I₀, and the final one, If:I₀ = 4 kg*m² + 1kg*(1.1 m)² + 1kg*(1.1m)² = 6.42 kg*m²  (3)If = 4 kg*m² + 1kg*(0.33m)² + 1kg*(0.33m)² = 4.22 kg*m² (4)Replacing (3), (4) in (1) we can solve for ωf:

       

b)

Since the student is not translating but he is only rotating, all his kinetic energy is rotational kinetic energy.The expression for the kinetic energy of a rotating rigid body, around an axis of rotation is as follows:

       

The initial kinetic energy of the student, before the objects are pulled in, is as follows:

     

The final kinetic energy is given by the following expression: