Question

On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbells at a distance of 0.620 m from the axis of rotation of the stool. She is given an angular velocity of 3.05 rad/s , after which she pulls the dumbbells in until they are only 0.160 m distant from the axis. The woman's moment of inertia about the axis of rotation is 5.50 kg⋅m2 and may be considered constant. Each dumbbell has a mass of 5.35 kg and may be considered a point mass. Neglect friction.

What is the angular velocity of the system after the dumbbells are pulled in toward the axis?

Compute the kinetic energy of the system before the dumbbells are pulled in.

Compute the kinetic energy of the system after the dumbbells are pulled in.

Answer 1

The initial moment of inertia of the woman dumbbell system
I = I_w + m r²
Where m is the mass of the dumbbell, m = 5.35 kg. I_w = 5.50 kg⋅m². r = 0.620 m, initially
I = 5.50 kg⋅m² + 5.35 kg x ( 0.620 m)²
I = 7.56 kg m²
The initial angular velocity _i = 3.05 rad/s
The angular momentum of the system, initially
L = I _i = 7.56 kg m² x 3.05 rad/s =  23.05 kg m²/s
This is equal to the final angular momentum, as the angular momentum is conserved.
The final moment of inertia of the system.
I = I_w + m r²
I = 5.50 kg⋅m² + 5.35 kg x ( 0.160 m)²
I = 5.64 kg m²
The final angular momentum
L_f = I _f
L_i = L_f
I _f =  23.05 kg m²/s
_f = 23.05 kg m²/s /  5.64 kg m²
_f = 4.09 rad/s
b) The final KE
KE = (1/2) I²
KE = (1/2) x 5.64 kg m² x (4.09 rad/s)²
KE =47.17 J