Answer to Question #118495 in Mechanics | Relativity for Wilcox Muchineripi

Question

What is the angular momentum of a figure skater spinning at 3.0 rev/s with arms in close to her body,
assuming her to be a uniform cylinder with a height of 1.5 m, a radius of 15 cm, and a mass of 48 kg ?
How much torque is required to slow her to a stop in 4.0 s, assuming she does not move her arms?

Answer 1

L=I$\omega=1/2 MR$²$\omega=1/2(48 kg)(0.15m)$² (3.0 rev/s) (2$\pi rad / 1 rev)=10 kg m^2/s$.

So, angular momentum of a figure skater spinning at 3.0 rev/s with arms in close to her body,

assuming her to be a uniform cylinder with a height of 1.5 m, a radius of 15 cm, and a mass of 48 kg is 10 kg m²/s.

If the rotational inertia does not change, then the charge in angular momentum is due to change in angular velocity.

$\tau=\Delta L / \Delta t$=0-10 / 4 = -1.2 Nm

Negative sign indicates that torque is in opposite direction of initial angular momentum.

 Torque of -1.2 Newton meters is required to slow her to a stop in 4.0 seconds.