Question #118372

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?

Calculate the torque required to slow her to a stop in 3.82s, assuming she does not move her arms?

Expert's answer

**Explanations & Calculations **

**1)**

- Angular velocity can be calculated from the given data,

"\\qquad\\qquad\n\\begin{aligned}\n\\small \\omega & = \\small 2\\pi f\\\\\n&= \\small 2\\pi\\times 3.06 s^{-1}\\\\\n&= \\small6.12\\pi s^{-1}\n\n\\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\n\\begin{aligned}\n\\small \\text{a.m }&= \\small I \\times \\omega\\\\\n&= \\small \\frac{1}{2}mR^2\\times \\omega\\\\\n&= \\small \\frac{1}{2}\\times 49.6kg \\times (15.2 \\times 10^{-2} m)^2\\times 6.12\\pi \\\\\n&= \\small \\bold{11.02kgm^2s^{-1}}\n\\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\n\\begin{aligned}\n\\small \\tau &= \\small \\frac{\\Delta I\\omega}{t} =\\frac{I\\Delta\\omega}{t}\\\\\n\\small -\\tau&= \\small \\frac{0.573kgm^2 \\times (0-6.12\\pi)}{3.82 s}\\\\\n\\small \\tau &= \\small \\bold{2.884 Nm}\n\\end{aligned}"

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