Question

A student sitting on a chair on a circular platform of negligible mass rotates freely on an air table at initial rotational speed 1.7 rad/s . The student's arms are initially extended with 6.0-kg dumbbells in each hand. As the student pulls her arms in toward her body, the dumbbells move from a distance of 0.80 m to 0.10 m from the axis of rotation. The initial rotational inertia of the student's body (not including the dumbbells) with arms extended is 6.0 kg⋅m2 and her final rotational inertia is 5.0 kg⋅m2 .

A) Determine the student's final rotational speed.

B) Determine the change of kinetic energy of the system consisting of the student together with the two dumbbells.

C) Determine the change in the kinetic energy of the system consisting of the two dumbbells alone without the student.

D) Determine the change of kinetic energy of the system consisting of student alone without the dumbbells.

Answer 1

(a) Conservation of angular momentum

I1 w1 = I2 w2 =

( 6 + 2 * 6 *0.8^2) * 2 = ( 5 + 2 * 6 * 0.1^2 ) w2

w2 = ( 6 + 2 * 6 *0.8^2) * 2 /( 5 + 2 * 6 * 0.1^2 ) = 5.343 rad/s

(b)  change of kinetic energy = KE2 - KE1 = 0.5 I2 w2^2 - 0.5 I1 w1^2

= 0.5 * ( 5 + 2 * 6 * 0.1^2 ) * 5.343^2 - 0.5*( 6 + 2 * 6 *0.8^2) * 2^2 = 45.72 J

(c) change of kinetic energy = 0.5 * ( 2 * 6 * 0.1^2 ) * 5.343^2 - 0.5*( 2 * 6 *0.8^2) * 2^2 = -13.65 J

(d) change of kinetic energy =  0.5 * ( 5 ) * 5.343^2 - 0.5*( 6 ) * 2^2 = 59.36 J