- 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?
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This is Giancoli Answers with Mr. Dychko. The angular momentum of the figure skater is their moment of inertia times their angular velocity. Moment of inertia for a cylinder is one-half mass times its radius squared and the height of the figure skater is not important, we only care about the radius or the distance from the axis of rotation which is the middle of her body to the outside of her body— maybe guess from her neck to her shoulder is probably this distance r here. So that's one-half times 48 kilograms times 0.15 meters squared times 3 revolutions per second written in radians per second by multiplying by 2π radians per revolution and we get 1.0 times 10 to the 1 kilogram meter squared per second. And I wrote this in scientific notation just to be clear that there are two significant figures; if I had just wrote the number 10 that would be one significant figure. So torque required to bring this figure skater to a stop is gonna be the rate of change of angular momentum so that's final angular momentum minus initial divided by time and there is no final angular momentum because they come to a stop and so it's negative L i over t. So that's negative 10.1788 kilogram meter squared per second divided by 4 seconds which is negative 2.5 newton meters of torque; the negative sign being the opposite to the initially positive angular velocity. So this torque is in the opposite direction to the initial angular velocity.
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