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Chapter 11: Problem 5

(II) A diver (such as the one shown in Fig. 2\()\) can reduce her moment of inertia by a factor of about 35 when changing from the straight position to the tuck position. If she makes 20 rotations in 1.5 \(\mathrm{s}\) when in the tuck position, what is her angular spced (rev/s) when in the straight position?

Short Answer

Expert verified
The angular speed in the straight position is approximately 0.38 rev/s.

Step by step solution

01

Understand the Given

The problem states that a diver reduces her moment of inertia by a factor of about 35 when changing from a straight to a tuck position. She makes 20 rotations in 1.5 seconds in the tuck position. We need her angular speed in the straight position.
02

Define the Variables and Formulas

Let \( I_s \) be the moment of inertia in the straight position and \( I_t \) in the tuck position. We know \( I_s = 35 \times I_t \). For angular speed \( \omega \), when the moment of inertia decreases, angular speed increases to conserve angular momentum: \( L = I \cdot \omega \). Hence \( I_s \cdot \omega_s = I_t \cdot \omega_t \).
03

Calculate Angular Speed in Tuck Position

The angular speed \( \omega_t \) in the tuck position is the number of rotations per second. Convert rotations to revolutions by dividing the number of rotations by the time: \( \omega_t = \frac{20}{1.5} \) rev/s.
04

Apply Conservation of Angular Momentum

Since angular momentum is conserved, set the initial and final angular momenta equal: \( I_s \cdot \omega_s = I_t \cdot \omega_t \). Rearrange to solve for \( \omega_s \): \( \omega_s = \frac{I_t \cdot \omega_t}{I_s} \).
05

Substitute Known Values

Substitute \( I_s = 35 \times I_t \) and \( \omega_t = \frac{20}{1.5} \) into the equation from Step 4: \( \omega_s = \frac{I_t \cdot \left(\frac{20}{1.5}\right)}{35 \times I_t} = \frac{20}{1.5 \times 35}\) rev/s.
06

Calculate Angular Speed in Straight Position

Simplify the expression from Step 5: \( \omega_s = \frac{20}{1.5 \times 35} \). Calculate to find \( \omega_s \approx 0.38 \) rev/s.

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Key Concepts

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

Conservation of Angular Momentum
The conservation of angular momentum is a fundamental principle in rotational motion, just like the conservation of linear momentum is in translational motion. It states that if no external torque acts on a system, its total angular momentum remains constant. This concept is pivotal in understanding why the diver can increase her spinning speed when moving into a tuck position.

In our specific scenario, the diver's angular momentum remains conserved as she transitions from a straight to a tuck position. Angular momentum, denoted as \( L \), is the product of the moment of inertia \( I \) and the angular speed \( \omega \) (\( L = I \cdot \omega \)).

  • When the diver pulls into a tuck position, she reduces her moment of inertia drastically, decreasing it by a factor of about 35 in this problem.
  • Since angular momentum must be conserved \( (I_s \cdot \omega_s = I_t \cdot \omega_t) \), a decrease in \( I \) results in an increase in \( \omega \), enabling the diver to spin faster.

If you find the conservation of angular momentum a bit tricky to grasp, imagine it like a pirouette in ice skating. When skaters pull their arms in, they spin faster because their moments of inertia decrease.
Angular Speed
Angular speed is a measure of how quickly an object rotates or spins around an axis. It tells us how many revolutions occur during a certain time period. In physics, angular speed is usually denoted by the Greek letter \( \omega \).

In the context of the diving exercise, we calculated the diver's angular speed in two different positions:
  • Tuck position: The diver makes 20 rotations in 1.5 seconds. Therefore, her angular speed in the tuck position \( \omega_t \) is \( \omega_t = \frac{20}{1.5} \) rev/s.
  • Straight position: In this position, her moment of inertia is much larger. By using the conservation of angular momentum, we found that her angular speed \( \omega_s \) is approximately 0.38 rev/s.

Angular speed is crucial as it illustrates how dynamic changes in body position can alter how fast rotations occur. Understanding this helps explain diverse movements in sports like gymnastics and diving.
Rotational Kinematics
Rotational kinematics, much like linear kinematics, deals with the motion of rotating objects. It involves parameters such as angular displacement, angular velocity, and angular acceleration.

  • Angular displacement refers to the angle through which an object rotates.
  • Angular velocity (or speed), \( \omega \), denotes how fast something spins.
  • Angular acceleration indicates how quickly the rotational speed changes.
In our diver's problem, although angular acceleration wasn't directly addressed, understanding its basics is helpful for further studies. The dive involves a significant change in angular speed as the diver moves between positions, illustrating principles of rotational kinematics.

Rotational movements are complex due to varying moments of inertia and changing body postures impacting rotational equations. Rotational kinematics helps decode these rotational behaviors by applying similar math and logic used in linear motion but adapted to circular paths.

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Most popular questions from this chapter

$$ \begin{array}{l}{\text { (1) Show that }(a) \hat{\mathbf{i}} \times \hat{\mathbf{i}}=\hat{\mathbf{j}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}} \times \hat{\mathbf{k}}=0,(b) \hat{\mathbf{i}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}},} \\ {\hat{\mathbf{i}} \times \hat{\mathbf{k}}=-\hat{\mathbf{j}}, \text { and } \hat{\mathbf{j}} \times \hat{\mathbf{k}}=\hat{\mathbf{i}}}\end{array} $$

Most of our Solar System's mass is contained in the Sun, and the planets possess almost all of the Solar System's angular momentum. This observation plays a key role in theories attempting to explain the formation of our Solar System. Estimate the fraction of the Solar System's total angular momentum that is possessed by planets using a simplified model which includes only the large outer planets with the most angular momentum. The central Sun (mass \(1.99 \times 10^{30} \mathrm{~kg}\), radius \(6.96 \times 10^{8} \mathrm{~m}\) ) spins about its axis once every 25 days and the planets Jupiter, Saturn, Uranus, and Neptune move in nearly circular orbits around the Sun with orbital data given in the Table below. Ignore each planet's spin about its own axis.

(1I) An Atwood machine (Fig. 16) consists of two masses, \(m_{\mathrm{A}}=7.0 \mathrm{kg}\) and \(m_{\mathrm{B}}=8.2 \mathrm{kg}\) . connected by a cord that passes over a pulley free to rotate about a fixed axis. The pulley is a solid cylinder of radius \(R_{0}=0.40 \mathrm{m}\) and mass 0.80 \(\mathrm{kg}\) . (a) Determine the acceleration \(a\) of each mass. (b) What percentage of error in \(a\) would be made if the moment of inertia of the pulley were ignored? Ignore friction in the pulley bearings.

(III) Suppose a 65 -kg person stands at the edge of a 6.5 -m dia- meter merry-go-round turntable that is mounted on frictionless bearings and has a moment of incrtia of 1850 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) . The tumtable is at rest initially, but when the person begins rumning at a speed of 3.8 \(\mathrm{m} / \mathrm{s}\) (with respect to the tumtablc) around its edge, the turntable begins to rotate in the opposite direction. Cakulate the angular velocity of the tumtable.

(II) A toy gyroscope consists of a \(170-\mathrm{g}\) disk with a radius of 5.5 \(\mathrm{cm}\) mounted at the center of a thin axle 21 \(\mathrm{cm}\) long (Fig. 41\()\) . The gyroscope spins at 45 \(\mathrm{rev} / \mathrm{s} .\) One end of its axle rests on a stand and the other end precesses horizontally about the stand. (a) How long does it take the gyroscope to precess once around? (b) If all the dimensions of the gyroscope were doubled (radius = \(11 \mathrm{cm},\) axle \(=42 \mathrm{cm} )\) how long would it take to precess once?
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