Question

What is the angular momentum of a figure skater spinning at 3.06 rev/s with arms in close to her body, assuming her body to be a uniform cylinder with a height of 1.55m, a radius of 15.2cm ,and a mass of 49.6kg?
Calculate the torque required to slow her to a stop in 3.82s, assuming she does not move her arms?

Answer 1

Explanations & Calculations

1)

• Angular velocity can be calculated from the given data,

$\qquad\qquad \begin{aligned} \small \omega & = \small 2\pi f\\ &= \small 2\pi\times 3.06 s^{-1}\\ &= \small6.12\pi s^{-1} \end{aligned}$

• Angular momentum is the product between angular velocity & the moment of inertia. Here she is assumed to be a solid cylinder .Therefore,

$\qquad\qquad \begin{aligned} \small \text{a.m }&= \small I \times \omega\\ &= \small \frac{1}{2}mR^2\times \omega\\ &= \small \frac{1}{2}\times 49.6kg \times (15.2 \times 10^{-2} m)^2\times 6.12\pi \\ &= \small \bold{11.02kgm^2s^{-1}} \end{aligned}$

2) As she is stopped due to the torque applied, there is a change in angular momentum during the period. By considering it, applied torque or the torque needed to stop her, can be found. Applying the equation along the spinning direction,

$\qquad\qquad \begin{aligned} \small \tau &= \small \frac{\Delta I\omega}{t} =\frac{I\Delta\omega}{t}\\ \small -\tau&= \small \frac{0.573kgm^2 \times (0-6.12\pi)}{3.82 s}\\ \small \tau &= \small \bold{2.884 Nm} \end{aligned}$