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Chapter 10: Problem 30

Why does a figure skater pull in her arms while increasing her angular velocity in a tight spin?

Short Answer

Expert verified
Answer: When a figure skater pulls in her arms during a spin, she decreases her moment of inertia by bringing her mass closer to the axis of rotation. Due to the conservation of angular momentum, this decrease in moment of inertia causes an increase in angular velocity, allowing her to spin more quickly.

Step by step solution

01

Understanding the conservation of angular momentum

The conservation of angular momentum states that the total angular momentum of a closed system remains constant if no external torques are acting on it. In our case, the closed system is the spinning figure skater. Since the skater is exerting no external torque on her body, her total angular momentum is conserved throughout the spin.
02

Angular momentum formula

Angular momentum (L) is the product of the moment of inertia (I) and angular velocity (ω). Mathematically, this can be represented as L = Iω. The moment of inertia depends on the mass distribution of the rotating object and the distance of the mass particles from the axis of rotation.
03

The effect of pulling in arms on the moment of inertia

When the figure skater pulls in her arms, she is effectively decreasing the distance of her mass particles from the axis of rotation. This reduction in distance decreases her moment of inertia.
04

The effect on angular velocity

Since angular momentum is conserved (L_initial = L_final), and the moment of inertia is decreased, the only way to maintain this conservation is by increasing the angular velocity. Mathematically, I_initialω_initial = I_finalω_final. If I_final is less than I_initial, ω_final must be greater than ω_initial.
05

Outcome

So, a figure skater pulls in her arms during a spin to decrease her moment of inertia, which, due to the conservation of angular momentum, results in an increase in her angular velocity. This increase in angular velocity is what allows the skater to spin more quickly in a tight spin.

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Most popular questions from this chapter

A uniform solid sphere of mass \(M\) and radius \(R\) is rolling without sliding along a level plane with a speed \(v=3.00 \mathrm{~m} / \mathrm{s}\) when it encounters a ramp that is at an angle \(\theta=23.0^{\circ}\) above the horizontal. Find the maximum distance that the sphere travels up the ramp in each case: a) The ramp is frictionless, so the sphere continues to rotate with its initial angular speed until it reaches its maximum height. b) The ramp provides enough friction to prevent the sphere from sliding, so both the linear and rotational motion stop (instantaneously).

The turbine and associated rotating parts of a jet engine have a total moment of inertia of \(25 \mathrm{~kg} \mathrm{~m}^{2}\). The turbine is accelerated uniformly from rest to an angular speed of \(150 \mathrm{rad} / \mathrm{s}\) in a time of \(25 \mathrm{~s}\). Find a) the angular acceleration, b) the net torque required, c) the angle turned through in \(25 \mathrm{~s}\) d) the work done by the net torque, and e) the kinetic energy of the turbine at the end of the \(25 \mathrm{~s}\).

Determine the moment of inertia for three children weighing \(60.0 \mathrm{lb}, 45.0 \mathrm{lb}\) and \(80.0 \mathrm{lb}\) sitting at different points on the edge of a rotating merry-go-round, which has a radius of \(12.0 \mathrm{ft}\).

A figure skater draws her arms in during a final spin. Since angular momentum is conserved, her angular velocity will increase. Is her rotational kinetic energy conserved during this process? If not, where does the extra energy come from or go to?

A thin uniform rod (length \(=1.00 \mathrm{~m},\) mass \(=2.00 \mathrm{~kg})\) is pivoted about a horizontal frictionless pin through one of its ends. The moment of inertia of the rod through this axis is \(\frac{1}{3} m L^{2} .\) The rod is released when it is \(60.0^{\circ}\) below the horizontal. What is the angular acceleration of the rod at the instant it is released?
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