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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

A small \(10.0 \mathrm{~g}\) bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass \(50.0 \mathrm{~g}\) and is \(100 \mathrm{~cm}\) in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of \(20.0 \mathrm{~cm} / \mathrm{s}\) relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps? (b) What is the total kinetic energy of the system just after the bug leaps? (c) Where does this energy come from?

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.

The moment of inertia of the empty turntable is \(1.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\). With a constant torque of \(2.5 \mathrm{~N} \cdot \mathrm{m},\) the turntable-person system takes \(3.0 \mathrm{~s}\) to spin from rest to an angular speed of \(1.0 \mathrm{rad} / \mathrm{s} .\) What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. (a) \(2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (b) \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (c) \(7.5 \mathrm{~kg} \cdot \mathrm{m}^{2} ;\) (d) \(9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\).

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

Example 10.7 discusses a uniform solid sphere rolling with- out slipping down a ramp that is at an angle \(\beta\) above the horizontal. Now consider the same sphere rolling without slipping up the ramp. (a) In terms of \(g\) and \(\beta\), calculate the acceleration of the center of mass of the sphere. Is your result larger or smaller than the acceleration when the sphere rolls down the ramp, or is it the same? (b) Calculate the friction force (in terms of \(M, g,\) and \(\beta\) ) for the sphere to roll without slipping as it moves up the incline. Is the result larger, smaller, or the same as the friction force required to prevent slipping as the sphere rolls down the incline?
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