Question

A student sits on a rotating stool holding two 3.2-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg · m2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.36 m from the rotation axis.

(a) Find the new angular speed of the student. rad/s

(b) Find the kinetic energy of the student before and after the objects are pulled in. before & after (J)

Answer 1

(a) This is a conservation of angular momentum problem.

Angular momentum = I * ω

For the stool + student, I = 3.0

Initial angular velocity = 0.75 rad/s

Initial angular momentum = 3 * 0.75 = 2.25

For each of the objects, I = m * r^2 = 3.2 * 1^2 = 3.2

For both objects, I =3.2*2 = 6.4

Initial angular velocity = 0.75 rad/s

Initial angular momentum = 6.4 * 0.75 = 4.8

Total initial angular momentum = 2.25 + 4.8 = 7.05

Let’s determine the momentum of the objects after they are pulled inward. r = 0.36 m

I = 6.4 * 0.36^2 = 0.82944

Angular momentum = 0.82944 * ωf

For the stool + student, angular momentum = 3 * ωf

Total final angular momentum = 3.82944 * ω

Total final angular momentum = Total initial angular momentum

3.82944 * ωf = 7.05

ωf = 1.84 rad/s

This is the new angular velocity of the student, stool, and 2 objects.

(b)

Rotational KE = ½ * I * ω^2

For the student and stool, initial KE = ½ * 3 * 0.75^2 = 0.84375

For the student and stool, final KE = ½ * 3 * (1.84)^2 = 5.08