Question

A professor sits on a rotating stool that is spinning at 10.0 rpm while she holds a heavy weight in each of her hands. Her outstretched hands are 0.765 m from the axis of rotation, which passes through her head into the center of the stool. When she symmetrically pulls the weights in closer to her body, her angular speed increases to 36.5 rpm. Neglecting the mass of the professor, how far are the weights from the rotational axis after she pulls her arms in?

Answer 1

L: angular momentum

Just as p = mv, L = Iω, I = sum of mr²

L₀ = 2mr₀²ω₀

Lf = 2mrf²ωf

we know r₀ = 0.765 m

ω₀ = 10 * 2π rad per minute, and ωf = 36.5 * 2π rad per minute

rf = r₀√(ω₀ / ωf) = 0.765 * √(10.0 / 36.5)

rf = 0.4004 m

Answer 2

Here angular momentum of the system remains conserved

$I_1\omega_1 = I_2\omega_2$

$mr_1^2\omega_1 = mr_2^2\omega_2$

0.765^2 (10) = r₂² (40.5)

r₂ = 0.38 m from rotational axis

Answer 3

Initial angular speed = $\omega$₁ = 10 rpm = 10 x (2$\pi$/60) rad/s = 1.0472 rad/s

Final angular speed = $\omega$₂ = 36.5 rpm = 36.5 x (2$\pi$/60) rad/s = 3.8222 rad/s

Mass of each heavy weight = M

There are 2 heavy weights as the professor holds one in each hand.

Initial distance of the weights from the axis of rotation = R₁ = 0.765 m

Final distance of the weights from the axis of rotation = R₂

By conservation of angular momentum,

(MR₁² + MR₁²)$\omega$₁ = (MR₂² + MR₂²)$\omega$₂

2MR₁²$\omega$₁ = 2MR₂²$\omega$₂

R₁²$\omega$₁ = R₂²$\omega$₂

(0.765)²(1.0472) = R₂²(3.8222)

R₂ = 0.4 m

Distance of the weights from the rotational axis after the professor pulls her arms in = 0.4 m