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Question

If anyone can help me out with this I would really appreciate it. I can not come up with the right answer that's in the book.

A student siits on a rotating stool holding two 4.0 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 x m2 and is assumed to be constant. The student then pulls the objects horizontally to 0.30 m from the rotation axis.

1. Find the new angular speed of the student. ____ rad/s

2. Find the kinetic energy of the student before and after the objects are pulled in. ____ J before and ____ J after

Answer 1

Before the total momentum of inertia is:
3[kg*m^2] + 2 *4[kg] * (1.0[m] ^2) = 11[kg*m^2]
After:
3[kg*m^2] + 2 *4[kg] * (0.3[m] ^2) = 3.72[kg*m^2]

Angular momentum(L) is constant so:
L=I*w
where L is angular momentum, I is momentum of inertia, w is angular velocity
so: 11[kg*m^2] * 0.75[rad/s] = 8.25[kg*m^2/s]
and: 8.25[kg*m^2/s] / 3.72[kg*m^2] = 2.218[rad/s]

Kinetic energy is: E = 0.5 * I * (w^2)
where E is kinetic energy, I is momentum of inertia, w is angular velocity.
Before: E1 = 0.5 * 11[kg*m^2] * (0.75[rad/s]^2) = 3.09375[J]
After: E2=0.5 * 3.72[kg*m^2] * (2.218[rad/s]^2) = 9.1503[J]

Answer 2

Inertial1 * omega 1 = Inertial2 * omega 2

The key to the problem is finding the formula for the inertial of of the 2 masses held in hands as a function of the radius from axis.

Inertial 1 = 2 masses in hands + student&stool#given#
omega1 = given

Calc Inertia2 and plug into equation...

sorry, too old to work out all the equations