/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 A diver can change his rotationa... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 78

A diver can change his rotational inertia by drawing his arms and legs close to his body in the tuck position. After he leaves the diving board (with some unknown angular velocity), he pulls himself into a ball as closely as possible and makes 2.00 complete rotations in \(1.33 \mathrm{s}\). If his rotational inertia decreases by a factor of 3.00 when he goes from the straight to the tuck position, what was his angular velocity when he left the diving board?

Short Answer

Expert verified
Answer: The diver's initial angular velocity when he left the diving board was 3.16 rad/s.

Step by step solution

01

Identify the knowns, unknowns, and constants

We are given: 1. The diver's rotational inertia decreases by a factor of 3.00 when he goes from the straight to the tuck position (let's call the straight position's rotational inertia \(I_1\), and the tuck position's rotational inertia \(I_2\)). 2. The diver makes 2.00 complete rotations in 1.33 s in the tuck position. 3. We need to find his initial angular velocity when he left the diving board (which we'll denote as \(\omega_1\)).
02

Express the given data in terms of equations

First, express the relationship between the diver's initial and final rotational inertia: \(I_2 = \frac{I_1}{3}\) Next, express the angular displacement (\(\theta\)) in terms of the number of rotations and time taken: \(\theta = 2.00 \times 2\pi\) Lastly, calculate the angular velocity (\(\omega_2\)) when he is in the tuck position: \(\omega_2 = \frac{\theta}{t}\), where \(t = 1.33 \mathrm{s}\)
03

Calculate the angular velocity in the tuck position

Using the equation \(\omega_2 = \frac{\theta}{t}\), we find: \(\omega_2 = \frac{2.00 \times 2\pi}{1.33 \mathrm{s}}\) \(\omega_2 = 9.47 \mathrm{rad/s}\)
04

Apply the conservation of angular momentum

Since there is no external torque acting on the diver, the angular momentum should be conserved: \(L_{initial} = L_{final}\) Expressing this in terms of rotational inertia and angular velocity: \(I_1 \omega_1 = I_2 \omega_2\) Using the equation \(I_2 = \frac{I_1}{3}\), we can rewrite the conservation of angular momentum equation as: \(I_1 \omega_1 = \frac{I_1}{3} \times \omega_2\)
05

Calculate the initial angular velocity

Now, we can solve for the initial angular velocity \(\omega_1\): \(\omega_1 = \frac{\omega_2}{3}\) Substitute the value of \(\omega_2\) that we calculated earlier: \(\omega_1 = \frac{9.47 \mathrm{rad/s}}{3}\) \(\omega_1 = 3.16 \mathrm{rad/s}\) So, the initial angular velocity when the diver left the diving board was \(\boxed{3.16 \mathrm{rad/s}}\).

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Most popular questions from this chapter

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A man is doing push-ups. He has a mass of \(68 \mathrm{kg}\) and his center of gravity is located at a horizontal distance of \(0.70 \mathrm{m}\) from his palms and \(1.00 \mathrm{m}\) from his feet. Find the forces exerted by the floor on his palms and feet.
A uniform diving board, of length \(5.0 \mathrm{m}\) and mass \(55 \mathrm{kg}\) is supported at two points; one support is located \(3.4 \mathrm{m}\) from the end of the board and the second is at \(4.6 \mathrm{m}\) from the end (see Fig. 8.19 ). What are the forces acting on the board due to the two supports when a diver of mass \(65 \mathrm{kg}\) stands at the end of the board over the water? Assume that these forces are vertical. (W) tutorial: plank) [Hint: In this problem, consider using two different torque equations about different rotation axes. This may help you determine the directions of the two forces.]
The string in a yo-yo is wound around an axle of radius \(0.500 \mathrm{cm} .\) The yo-yo has both rotational and translational motion, like a rolling object, and has mass \(0.200 \mathrm{kg}\) and outer radius \(2.00 \mathrm{cm} .\) Starting from rest, it rotates and falls a distance of \(1.00 \mathrm{m}\) (the length of the string). Assume for simplicity that the yo-yo is a uniform circular disk and that the string is thin compared to the radius of the axle. (a) What is the speed of the yo-yo when it reaches the distance of $1.00 \mathrm{m} ?$ (b) How long does it take to fall? [Hint: The transitional and rotational kinetic energies are related, but the yo-yo is not rolling on its outer radius.]
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