Question

A 45 kg figure skater is spinning on the toes of her skates at 1.0 rev/s . Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 kg , 20 cmaverage diameter, 160 cm tall) plus two rod-like arms (2.5 kg each, 64 cm long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 kg, 20-cm-diameter, 200-cm-tall cylinder.

What is her new rotation frequency, in revolutions per second?

w2 = __________

Answer 1

The concept you're dealing with here is: Conservation of Angular Momentum (same as conservation of momentum, but in the rotational world). Angular Momentum (L) can be found using: L(final) = L(initial) I * w (final) = I * w (initial) Where I is the moment of inertia of the object and w is the angular velocity of the object. The trick here is finding her moment of inertia in each case: Case 1: Her bod is a cylinder spinning around its central axis: radius = 0.1 m, mass = 40kg. The moment of inertia for a cylinder is: I = m * r^2 / 2 ==> I = 0.2 [kg * m^2] Her arms are like rods: mass = 2.5kg, length = 0.66m.you can find the moment of inertia of a rod rotated about its perpendicular axis: I = m * L^2 / 12 ==> I = 0.09075 [kg * m^2] But her arms aren't rotating about their perpendicular axis, they are rotating about the center of her body which is some distance away. That distance can be expressed as half the length of her arm plus the radius of her body: D = 0.66m / 2 + 0.1 ==> D = 0.43 m Using the parallel axis theorem, you can now find the moment of inertia of each arm: I = I(center) + m * D^2 I = 0.09075 + 2.5 * 0.43^2 ==> I = 0.553 [kg * m^2] This is the moment of inertia for one arm. We already have the moment of inertia for her body, so we can add the body plus two arms to get her total moment of inertia: I = I(body) + 2 * I(arms) I = 0.2 + 2 * (0.553) == > I = 1.306 [kg * m^2] So, using the equation for angular momentum, we know her initial momentum is: L(initial) = I * w L = 1.306 [kg * m^2] * 1 [rev/s] NOTE: I didn't change the units of the angular speed because these are the units we want to find. If I converted it here to rad/sec, I would have to convert it back to rev/s later. Just a quick shortcut. Now, using conservation of momentum, we know the final momentum must be equal to the original momentum. So, first we need to find her new moment of inertia. Case 2: She is now one big cylinder: radius = 0.1 m, mass = 45kg: I = m * r^2 / 2 ==> I = 0.225 [kg * m^2]. And finally: L(final) = L(initial) I * w (final) = 1.306 (from before) w = 1.306 / I w = 1.306 / 0.225 ==> w = 5.804444... rev/s Rounded to two sig figs: w = 5.80 rev/s