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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 \(\mathrm{s}\) s. If she hadn't tucked at all, how many revolutions would she have made in the 1.5 s from board to water?

Short Answer

Expert verified
0.6 revolutions.

Step by step solution

01

Calculate Angular Velocity While Tucked

When the diver is tucked, she makes two complete revolutions in 1.0 s. One revolution is equal to \(2\pi\) radians. Therefore, in 1 second, she covers \(2 \times 2\pi = 4\pi\) radians. Her angular velocity \(\omega_2\) while tucked is then given by \(\omega_2 = \frac{\text{total radians}}{\text{time}} = \frac{4\pi}{1.0} = 4\pi \text{ rad/s}\).
02

Apply Conservation of Angular Momentum

The conservation of angular momentum states that the initial angular momentum must equal the final angular momentum. This can be written as:\[I_1 \omega_1 = I_2 \omega_2\]where \(I_1\) is the initial moment of inertia (18 \(\mathrm{kg} \cdot \mathrm{m}^{2}\)) and \(I_2\) is the moment of inertia while tucked (3.6 \(\mathrm{kg} \cdot \mathrm{m}^{2}\)). We need to find \(\omega_1\). Rearrange for \(\omega_1\):\[\omega_1 = \frac{I_2 \omega_2}{I_1} = \frac{3.6 \times 4\pi}{18} = \frac{14.4\pi}{18} = 0.8\pi \text{ rad/s}\]
03

Calculate Revolutions Without Tucking

Now that we have \(\omega_1 = 0.8\pi\) rad/s, calculate the number of revolutions in 1.5 seconds:\[\text{Revolutions} = \frac{\text{Total angle in rad}}{2\pi} = \frac{0.8\pi \times 1.5}{2\pi} = \frac{1.2\pi}{2\pi} = 0.6\]So she would have made 0.6 revolutions if she hadn't tucked.

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Key Concepts

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

Moment of Inertia
The concept of 'moment of inertia' plays a crucial role in understanding rotational motion, akin to mass in linear motion. To put it simply, the moment of inertia is a measure of how difficult it is to change an object's rotational state. It's important to note that moment of inertia depends not only on the object's mass but also on how that mass is distributed relative to the axis of rotation.

In our diver example, this is vividly illustrated when she changes her body's configuration. Initially, when her limbs are extended, the moment of inertia is high at 18 kg路m虏. Tucking into a ball decreases her moment of inertia to 3.6 kg路m虏, significantly reducing the resistance against rotational acceleration.
  • A large moment of inertia means more "rotational weight," making it harder to accelerate or decelerate.
  • Adjusting body shape can drastically alter an object's rotational properties.
This relationship exemplifies why divers tuck during a dive to spin faster鈥攕maller moments of inertia allow for greater angular velocities.
Angular Velocity
Angular velocity describes how fast an object rotates.It is the rotational equivalent of linear speed and expressed in units of radians per second (rad/s).One full rotation is equal to 2蟺 radians. Understanding angular velocity helps clarify how quickly a diver completes multiple rotations.

In the exercise, when the diver tucks, she accelerates due to her decreased moment of inertia, achieving an angular velocity of 4蟺 rad/s.This means she can complete two full turns (or revolutions) in just one second.
  • Angular velocity formula: \( \omega = \frac{\text{Total angle in rad}}{\text{time in s}} \).
  • It's a critical factor in sports and machinery where rotational speed is of the essence.
In practical terms, angular velocity dictates how fast athletes can spin, allowing divers and gymnasts to execute intricate turns and twists with precision. For our diver, not tucking means slower spins at 0.8蟺 rad/s, equating to less than one full revolution in the same time frame.
Rotational Motion
Rotational motion captures how objects turn about an axis and integrates concepts like moment of inertia and angular velocity. The study of rotational motion is invaluable, given its application in multiple fields ranging from mechanical systems to athletic movements.

The diver's scenario offers a classic application of rotational dynamics principles. As she tucks, she experiences a quicker spin due to the conserved angular momentum, a key law in rotational motion, which ensures that the product of the moment of inertia and angular velocity remains constant unless acted upon by an external torque.
  • Rotational motion involves considerations of torque, rotational kinematics, and energy conservation.
  • It can be utilized in designing efficient mechanical parts like gears and turbines.
By manipulating her moment of inertia through body position, the diver exemplifies fundamental physics at play in sports. In summary, rotational motion isn't just theoretical; it's central to diverse practical scenarios including vehicle dynamics and even planetary rotations.

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Most popular questions from this chapter

A 55-kg runner runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 2.8 \(\mathrm{m} / \mathrm{s}\) . The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.20 rad/s relative to the earth. The radius of the turntable is \(3.0 \mathrm{m},\) and its moment of inertia about the axis of rotation is 80 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can model the runner as a particle.)

A Gyroscope on the Moon. A certain gyroscope precesses at a rate of 0.50 \(\mathrm{rad} / \mathrm{s}\) when used on earth. If it were taken to a lunar base, where the acceleration due to gravity is \(0.165 g,\) what would be its precession rate?

A uniform solid cylinder with mass \(M\) and radius 2\(R\) rests on a horizontal tabletop. A string is attached by a yoke to a frictionless axle through the center of the cylinder so that the cylinder can rotate about the axle. The string runs over a disk-shaped pulley with mass \(M\) and radius \(R\) that is mounted on a frictionless axle through its center. A block of mass \(M\) is suspended from the free end of the string (Fig. P10.87). The string doesn't slip over the pulley surface, and the cylinder rolls without slipping on the tabletop. Find the magnitude of the acceleration of the block after the system is released from rest.

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 \(R / 2\) .

One force acting on a machine part is \(\vec{\boldsymbol{F}}=(-5.00 \mathrm{N}) \hat{\imath}+\) \((4.00 \mathrm{N}) \hat{\boldsymbol{J}}\) . The vector from the origin to the point where the force is applied is \(\vec{r}=(-0.450 \mathrm{m}) \hat{\imath}+(0.150 \mathrm{m}) \hat{\boldsymbol{J}}\) (a) In a sketch, show \(\vec{r}, \vec{\boldsymbol{F}},\) and the origin. (b) Use the right-hand rule to determine the direction of the torque. (c) Calculate the vector torque for an axis at the origin produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).
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