A figure skater can be idealized as composed of two solid cylinders of uniform density crossed at right angles to one another. Assume that the skater is spinning with angular speed \(\omega\) around a vertical axis passing through the center of her body. The mass of the large cylinder is \(M\); it has radius \(R\) and length \(L\). The mass of the small cylinder is \(m\); its radius is \(r\) and its length is \(\ell\). a) What is the moment of inertia of the large vertical cylinder in terms of the given quantities? b) What is the moment of inertia of the horizontal cylinder in terms of the given quantities? c) What is the figure skater's initial angular momentum? d) If \(m=0.1 M, \ell=8 R\) and the skater's arms can be totally retracted to her body, what is her new angular speed in terms of \(\omega\) ?

Short Answer

Expert verified
In order to calculate the moment of inertia for the large vertical cylinder, we must first identify the mass (M) and radius (R) of the cylinder. Let's assume the mass of the large vertical cylinder is M1 and its radius is R1. Then, the moment of inertia for the large vertical cylinder (I1) can be calculated using the formula: I1 = (1/2)M1R1^2

Step by step solution

01

Calculate moment of inertia for the large vertical cylinder

For a solid cylinder rotating about a central axis, its moment of inertia I is given by the formula: \(I = \frac{1}{2}MR^2\). This is the moment of inertia of the large vertical cylinder.

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