Question

I have part 1 done and correct.. it is 0.9741571435, i can’t get part 2 done!

Question (part 1 of 2) 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 iner- tia 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. Calculate the final angular speed of the student. Answer in units of rad/s.

Answer 1

here,

the moment of inertia of stool and student , I = 4 kg.m^2

r1 = 1.1 m , r2 = 0.33 m

mass of each object , m = 1 kg

the initial angular speed , w0 = 0.64 rad/s

let the final angular speed be w

using conservation of angular momentum

(I + 2 * m * r1^2) * w0 = ( I + 2 * m * r2^2) * w

( 4 + 2 * 1 * 1.1^2) * 0.64 = ( 4 + 2 * 1 * 0.33^2) * w

solving for w

w = 0.97 rad/s

the final angular speed is 0.97 rad/s

the change in kinetic energy , dKE = 0.5 * ( ( I + 2 * m * r2^2) * w^2 - (I + 2 * m * r1^2) * w0^2 )

dKE = 0.5 * ( ( 4 + 2 * 1 * 0.33^2) * 0.97^2 - ( 4 + 2 * 1 * 1.1^2) * 0.64^2)

dKE = 0.67 J