A student sits on a rotating stool holding two 5 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 0.73 rad/sec. The moment of inertia of the student plus the stool is 2 kg m^2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.31 m from the rotation axis.
(a) Calculate the final angular speed of the student. Answer in units of rad/s.
(b) Find the kinetic energy of the student before and after the objects are pulled in.

Explanation:

Given that,

Two object of Mass = 5kg

M1 = M2 = 5kg

Radius of rotation r = 1.1m

Angular speed wi = 0.73rad/s

Moment of inertia I = 2kgm²

The student pull the mass to another position of r' = 0.31m

A. Final angular momentum?

Using the conservation of angular momentum

The momentum of inertia of the objects the students holding I = mr²

L(initial) = L(final)

I•wi = I•wf

(I+mr²+mr²)•wi = (I+mr'²+mr'²)•wf

(2+5•1.1²+5•1.1²)•0.73=(2+5•0.31²+5•0.31²)wf

10.293 = 2.961wf

Then, wf= 10.293/2.961

wf = 3.476 rad/s

wf ≈ 3.48rad/s

b) kinetic energy before and after

Kinetic energy is give as

Kinetic energy = ½•I•w²

Initial kinetic energy is

K.E(initial) = ½(I+mr²+mr²)wi²

K.E(initial) = ½(2+5•1.1²+5•1.1²)•0.73²

K.E(initial) = 3.76 J

Final kinetic energy is

K.E(final) = ½(I+mr'²+mr'²)wf²

K.E(final)=½(2+5•0.31²+5•0.31²)3.48²

K.E(final) = 17.93J

answer is slate

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