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Chapter 8: Problem 74

A figure skater is spinning at a rate of 1.0 rev/s with her arms outstretched. She then draws her arms in to her chest, reducing her rotational inertia to \(67 \%\) of its original value. What is her new rate of rotation?

Short Answer

Expert verified
Answer: The new rate of rotation is approximately 1.49 rev/s.

Step by step solution

01

Write the conservation of angular momentum equation

Before and after the skater draws her arms in, the conservation of angular momentum equation states that: \( I_i \omega_i = I_f \omega_f \), where \(I_i\) is the initial moment of inertia, \(\omega_i\) is the initial angular velocity, \(I_f\) is the final moment of inertia, and \(\omega_f\) is the final angular velocity.
02

Find the final moment of inertia

According to the problem, the final moment of inertia (\(I_f\)) is 67% of the initial moment of inertia (\(I_i\)). Therefore, \(I_f = 0.67 I_i\)
03

Substitute the known values into the conservation of angular momentum equation

We know the initial angular velocity (\(\omega_i\)) is 1.0 rev/s, so we can rewrite the equation as: \(I_i(1\,\text{rev/s}) = (0.67 I_i) \omega_f\)
04

Solve for the final angular velocity (\(\omega_f\))

We can now solve for the final angular velocity (\(\omega_f\)) by dividing both sides of the equation by \(0.67I_i\): \(\omega_f = \dfrac{I_i(1\,\text{rev/s})}{0.67I_i}\)
05

Simplify and calculate the final angular velocity

Therefore, the final angular velocity is: \(\omega_f = \dfrac{1\,\text{rev/s}}{0.67} \approx 1.49\,\text{rev/s}\) So, when the figure skater draws her arms in to her chest, her new rate of rotation is approximately 1.49 rev/s.

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Most popular questions from this chapter

A skater is initially spinning at a rate of 10.0 rad/s with a rotational inertia of \(2.50 \mathrm{kg} \cdot \mathrm{m}^{2}\) when her arms are extended. What is her angular velocity after she pulls her arms in and reduces her rotational inertia to \(1.60 \mathrm{kg} \cdot \mathrm{m}^{2} ?\)
A 1.10 -kg bucket is tied to a rope that is wrapped around a pole mounted horizontally on friction-less bearings. The cylindrical pole has a diameter of \(0.340 \mathrm{m}\) and a mass of \(2.60 \mathrm{kg} .\) When the bucket is released from rest, how long will it take to fall to the bottom of the well, a distance of \(17.0 \mathrm{m} ?\)
A child of mass \(40.0 \mathrm{kg}\) is sitting on a horizontal seesaw at a distance of \(2.0 \mathrm{m}\) from the supporting axis. What is the magnitude of the torque about the axis due to the weight of the child?
What is the rotational inertia of a solid iron disk of mass \(49 \mathrm{kg},\) with a thickness of \(5.00 \mathrm{cm}\) and radius of \(20.0 \mathrm{cm}\) about an axis through its center and perpendicular to it?
The distance from the center of the breastbone to a man's hand, with the arm outstretched and horizontal to the floor, is \(1.0 \mathrm{m} .\) The man is holding a 10.0 -kg dumbbell, oriented vertically, in his hand, with the arm horizontal. What is the torque due to this weight about a horizontal axis through the breastbone perpendicular to his chest?
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