A figure skater can increase her spin rotation rate from an...

A figure skater can increase her spin rotation rate from an initial rate of 1.0 rev every $1.5 \mathrm{~s}$ to a final rate of $2.5 \mathrm{rev} / \mathrm{s} .$ If her initial moment of inertia was $4.6 \mathrm{~kg} \cdot \mathrm{m}^{2}$ what is her final moment of inertia? How does she physically accomplish this change?

Key Concepts

Conservation of Angular Momentum

This concept states that if no external torques act on a system, its angular momentum remains constant over time. In the context of a figure skater, as she pulls her arms inward, she reduces her moment of inertia, and to conserve angular momentum, her rotational speed increases.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotation, determined by both its mass and the distribution of that mass relative to the axis of rotation. By bringing her arms closer to her body, the figure skater decreases the distance of her mass from the axis, leading to a lower moment of inertia.

Effect of Body Configuration on Rotation

Changing the distribution of mass in a rotating body directly affects its spin rate. In scenarios like figure skating, altering the body's configuration—such as pulling in the arms—allows the skater to control and adjust her spin speed by modifying her moment of inertia.

Transcript

00:01 So in this problem we are given that a figure skater is rotating about her own axis, central axis.

00:08 And we are given that it originally she has a omega or rotational speed of one revolution per every 1 .5 seconds.

00:19 And finally she has a omega of 2 .5 revolutions per second.

00:28 And we are told that she has initially a moment of inertia of 4 .6.

00:34 Kg meter square and we are supposed to find the moment of inertia of the second case.

00:42 So in this case we will assume that there is no external torque that helps the figure skater to go from omega 1 to omega 2.

00:51 We assume that she is doing it on her own internally such that we can say that the original angular momentum will be same as the final angular momentum of the system and this is true for any system which is rotating but has no external torque to disturb the angular momentum so we'll use this and so before we do that let's convert the omegas into a common unit so this is one revolution per 1 .5 seconds or if you divide by 1 .5 you get 0 .67 revolutions per second and this in radiance per second if you multiply by 2 pi you get 4 .2 radiance per second so that's the first conversion and second conversion and second conversion is a bit easier, it is already in rotation, rotations per or revolutions per second.

01:41 You can multiply this by 2 pi because one revolution is equal to 2 pi radiance.

01:45 So we say this becomes equal to 15 .7 radiance per second.

01:51 So you see the difference is huge here, 4 .2 and 15 .7, the common units.

01:56 All right, so now we can go and take a look at this equation...

Please give Ace some feedback