Question

A student on a piano stool rotates freely with an angular speed of 3.07 rev/s. The student holds a 1.44-kg mass in each outstretched arm, 0.785 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.39 kg$\cdot$m2, a value that remains constant. As the student pulls his arms inward, his angular speed increases to 3.48 rev/s. How far are the masses from the axis of rotation at this time, considering the masses to be points?

Answer 1

Use conservation of angular momentum: L = Iω is constant. Let I0 = 5.39 kg·m^2, m = 1.44 kg, r_i = 0.785 m. Initial and final angular speeds (rad/s):
ω_i = 3.07(2π) = 19.2894 s^-1,
ω_f = 3.48(2π) = 21.8655 s^-1.

I_initial = I0 + 2 m r_i^2 = 5.39 + 2(1.44)(0.785^2) = 7.1647 kg·m^2.

Conserve L:
(I0 + 2 m r_i^2) ω_i = (I0 + 2 m r_f^2) ω_f

Solve for r_f^2:
r_f^2 = [ (I0 + 2 m r_i^2) ω_i - I0 ω_f ] / (2 m ω_f)

Plugging in:
r_f^2 ≈ 20.3482 / 62.9872 ≈ 0.3233 m^2

r_f ≈ sqrt(0.3233) ≈ 0.569 m.

So each mass is about 0.569 m from the axis.