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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 (蠅). Mathematically, this can be represented as L = I蠅. 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_initial蠅_initial = I_final蠅_final. If I_final is less than I_initial, 蠅_final must be greater than 蠅_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

In a tire-throwing competition, a man holding a \(23.5-\mathrm{kg}\) car tire quickly swings the tire through three full turns and releases it, much like a discus thrower. The tire starts from rest and is then accelerated in a circular path. The orbital radius \(r\) for the tire's center of mass is \(1.10 \mathrm{~m},\) and the path is horizontal to the ground. The figure shows a top view of the tire's circular path, and the dot at the center marks the rotation axis. The man applies a constant torque of \(20.0 \mathrm{~N} \mathrm{~m}\) to accelerate the tire at a constant angular acceleration. Assume that all of the tire's mass is at a radius \(R=0.35 \mathrm{~m}\) from its center. a) What is the time, \(t_{\text {throw }}\) required for the tire to complete three full revolutions? b) What is the final linear speed of the tire's center of mass (after three full revolutions)? c) If, instead of assuming that all of the mass of the tire is at a distance \(0.35 \mathrm{~m}\) from its center, you treat the tire as a hollow disk of inner radius \(0.30 \mathrm{~m}\) and outer radius \(0.40 \mathrm{~m}\), how does this change your answers to parts (a) and (b)?

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