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Chapter 10: Problem 44

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of \(18 \mathrm{~kg} \cdot \mathrm{m}^{2}\). She then tucks into a small ball, decreasing this moment of inertia to \(3.6 \mathrm{~kg} \cdot \mathrm{m}^{2}\). While tucked, she makes two complete revolutions in \(1.0 \mathrm{~s}\). If she hadn't tucked at all, how many revolutions would she have made in the \(1.5 \mathrm{~s}\) from board to water?

Short Answer

Expert verified
If the diver hadn't tucked at all, she would have made 0.6 revolutions in 1.5 seconds from board to water

Step by step solution

01

Find the initial angular velocity

Let's denote the initial moment of inertia as \(I_1 = 18 \, \mathrm{kg} \cdot \mathrm{m}^2\), and the final moment of inertia when tucked as \(I_2 = 3.6 \, \mathrm{kg} \cdot \mathrm{m}^2\). The number of revolutions per second (angular velocity) when tucked can be denoted as \(\omega_2 = 2 \, \mathrm{rev/s}\). In order to find the initial angular velocity, \(\omega_1\), we can use the conservation of angular momentum, which states that \(I_1\omega_1 = I_2\omega_2\). Solving for \(\omega_1\) gives us \(\omega_1 = \frac{I_2\omega_2}{I_1}\)
02

Calculate the initial angular velocity

Plugging in the given values: \(\omega_1 = \frac{(3.6 \, \mathrm{kg} \cdot \mathrm{m}^2)(2 \, \mathrm{rev/s})}{18 \, \mathrm{kg} \cdot \mathrm{m}^2} = 0.4 \, \mathrm{rev/s}\)
03

Step 3:Determine the number of revolutions

The number of revolutions the diver would have made in the 1.5 s from board to water if she hadn't tucked at all can be calculated by multiplying the time of descent (1.5 s) by the initial angular velocity. Number of revolutions = \(\omega_1 \cdot t = 0.4 \, rev/s \cdot 1.5 \, s = 0.6 \, revolutions\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
Moment of inertia is like the rotational equivalent of mass in linear motion. It tells us how difficult it is to change the rotation of an object. For example, a diver with straight arms and legs has a large moment of inertia because her body is spread out from her axis of rotation. This means it is hard for her to spin quickly.

When she tucks into a small ball, her body is closer to her axis of rotation, reducing her moment of inertia. This reduction makes it easier for her to increase her angular velocity, or spin faster. Think of it like spinning in a swivel chair: when you bring your arms in, you spin faster.

The diver's change in body position from straight to tucked drastically lowers her moment of inertia from \(18 \, \mathrm{kg} \cdot \mathrm{m}^2\) to \(3.6 \, \mathrm{kg} \cdot \mathrm{m}^2\). This plays a crucial role in how she changes her speed of rotation, which is governed by the conservation of angular momentum.
Angular Velocity
Angular velocity describes how fast an object spins or rotates. It is measured in revolutions per second (rev/s) or in other units like radians per second. Just like linear velocity tells us how fast something moves in a straight line, angular velocity indicates how fast something is spinning.

For the diver, her angular velocity changes when she transitions from a straight posture to a tucked position. Initially, with her arms outstretched, she has a lower angular velocity because of the higher moment of inertia. As she tucks in, her moment of inertia decreases significantly, allowing her angular velocity to increase. This increase lets her spin faster, performing more revolutions in the same amount of time.

In the exercise, while tucked, the diver spins at \(2 \, \mathrm{rev/s}\). By using the conservation of angular momentum \((I_1 \omega_1 = I_2 \omega_2)\), we calculated her initial angular velocity to be \(0.4 \, \mathrm{rev/s}\). This demonstrates how her body's configuration affects her spinning speed.
Revolutions
Revolutions quantify the number of complete turns or spins an object makes. Simply put, if you spin once completely around, that's one revolution.

In the context of the diver example, revolutions help us understand how much she spins during her dive from the board to the water. Initially, if she had not tucked in, she would have completed \(0.6\) revolutions in \(1.5\) seconds. This calculation is straightforward by multiplying the initial angular velocity \( (0.4 \, \mathrm{rev/s}) \) by the total time she falls \((1.5 \, \mathrm{s})\).

Understanding revolutions is essential as it provides insight into the effectiveness of the diver's body positioning and how changing the moment of inertia allows for more or fewer turns in mid-air. In many sports and applications, knowing how many revolutions an object makes can help fine-tune practices and performances, making revolutions a key metric in rotational dynamics.

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Most popular questions from this chapter

The Atwood's Machine. Figure P10.63 illustrates an Atwood's machine. Find the linear accelerations of blocks \(A\) and \(B,\) the angular acceleration of the wheel \(C,\) and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. Let the masses of blocks \(A\) and \(B\) be 4.00 \(\mathrm{kg}\) and \(2.00 \mathrm{~kg},\) respectively, the moment of inertia of the wheel about its axis be \(0.220 \mathrm{~kg} \cdot \mathrm{m}^{2}\), and the radius of the wheel be \(0.120 \mathrm{~m}\).

A \(12.0 \mathrm{~kg}\) box resting on a horizontal, frictionless surface is attached to a \(5.00 \mathrm{~kg}\) weight by a thin, light wire that passes over a frictionless pulley (Fig. E10.16). The pulley has the shape of a uniform solid disk of mass \(2.00 \mathrm{~kg}\) and diameter \(0.500 \mathrm{~m} .\) After the system is released, find (a) the tension in the wire on both sides of the pulley, (b) the acceleration of the box, and (c) the horizontal and vertical components of the force that the axle exerts on the pulley.

In your job as a mechanical engineer you are designing a flywheel and clutch- plate system like the one in Example \(10.11 .\) Disk \(A\) is made of a lighter material than disk \(B\), and the moment of inertia of disk \(A\) about the shaft is one-third that of disk \(B .\) The moment of inertia of the shaft is negligible. With the clutch disconnected, \(A\) is brought up to an angular speed \(\omega_{0} ; B\) is initially at rest. The accelerating torque is then removed from \(A,\) and \(A\) is coupled to \(B\). (Ignore bearing friction.) The design specifications allow for a maximum of \(2400 \mathrm{~J}\) of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk \(A\) so as not to exceed the maximum allowed value of the thermal energy?

What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? (a) A uniform solid cylinder; (b) a uniform sphere; (c) a thin-walled, hollow sphere; (d) a hollow cylinder with outer radius \(R\) and inner radius \(R / 2\).

If the body's center of mass were not placed on the rotational axis of the turntable, how would the person's measured moment of inertia compare to the moment of inertia for rotation about the center of mass? (a) The measured moment of inertia would be too large; (b) the measured moment of inertia would be too small; (c) the two moments of inertia would be the same; (d) it depends on where the body's center of mass is placed relative to the center of the turntable.
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