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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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