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Chapter 11: Problem 68

A top spins at \(30 \mathrm{rev} / \mathrm{s}\) about an axis that makes an angle of \(30^{\circ}\) with the vertical. The mass of the top is \(0.50 \mathrm{~kg}\), its rotational inertia about its central axis is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and its center of mass is \(4.0 \mathrm{~cm}\) from the pivot point. If the spin is clockwise from an overhead view, what are the (a) precession rate and (b) direction of the precession as viewed from overhead?

Short Answer

Expert verified
(a) 1.04 rad/s, (b) counterclockwise

Step by step solution

01

Find the Angular Speed

First, convert the angular speed from revolutions per second to radians per second. Since 1 revolution equals \(2\pi\) radians, the angular speed \(\omega\) is given by: \[ \omega = 30 \times 2\pi = 60\pi \text{ rad/s} \]
02

Determine the Torque

Torque (\(\tau\)) about the pivot point is created by the gravitational force acting at the center of mass. Use the formula:\[ \tau = r \cdot F \cdot \sin(\theta) \]where \(r = 0.04 \text{ m} \) is the distance from the pivot to the center of mass, \( F = m \cdot g = 0.50 \times 9.8 = 4.9 \text{ N} \), and \( \theta = 30^{\circ} \). Thus, \[ \tau = 0.04 \cdot 4.9 \cdot \sin(30^{\circ}) = 0.04 \cdot 4.9 \cdot 0.5 = 0.098 \text{ Nm} \]
03

Calculate the Precession Rate

The precession rate \( \Omega \) is given by:\[ \Omega = \frac{\tau}{I \cdot \omega} \]where \( I = 5.0 \times 10^{-4} \text{ kg}\cdot\text{m}^2 \) is the rotational inertia and \( \omega = 60\pi \text{ rad/s} \) from Step 1. Therefore,\[ \Omega = \frac{0.098}{5.0 \times 10^{-4} \cdot 60\pi} = \frac{0.098}{0.09425} \approx 1.04 \text{ rad/s} \]
04

Identify the Direction of Precession

Since the top spins clockwise from an overhead view and the torque due to gravity acts perpendicular to the axle, the precession will occur counterclockwise from above. This is due to the right-hand rule, which suggests that the torque vector (up) causes a counterclockwise precession.

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Key Concepts

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

Angular Speed
Angular speed measures how fast something spins around a central point or axis. It tells us the distance covered in radians per second, which is crucial for understanding rotational movements. In this exercise, the top spins at 30 revolutions per second. To convert revolutions into a more standardized unit, we use radians. Since 1 revolution equals \(2\pi\) radians, the angular speed \(\omega\) of the top becomes:
  • \(\omega = 30 \times 2\pi = 60\pi \text{ rad/s}\)
This conversion helps relate the top's rotational motion to other factors like torque and precession rate. Angular speed is foundational when calculating how other rotational dynamics evolve, such as the rate at which a spinning object precesses.
Torque Calculation
Torque is a force that causes objects to rotate around an axis. It depends on three main factors: the strength of the force applied, the distance from the pivot point, and the angle at which the force is applied. The formula for torque \(\tau\) is:
  • \[ \tau = r \cdot F \cdot \sin(\theta) \]
In this exercise, torque results from the center of mass of the top acting through gravity. Let's break down each part:
  • \(r = 0.04 \text{ m}\) is the distance from the pivot to the center of mass.
  • \(F = m \cdot g = 0.50 \times 9.8 = 4.9 \text{ N}\) is the force due to gravity.
  • \(\theta = 30^{\circ}\), so \(\sin(30^{\circ}) = 0.5\).
  • Hence, \(\tau = 0.04 \cdot 4.9 \cdot 0.5 = 0.098 \text{ Nm}\)
Understanding torque goes hand in hand with concepts like rotational inertia and precession calculations.
Rotational Inertia
Rotational inertia, also known as the moment of inertia, measures an object's resistance to changes in its rotation. It depends on an object's mass and how that mass is distributed in relation to the axis of rotation.
The formula is typically denoted as \(I\) and varies based on the object's shape and axis.
In this exercise, the top has a central axis rotational inertia of \(5.0 \times 10^{-4} \text{ kg} \cdot \text{m}^2\).
  • This value tells us how much torque will be needed to alter the spinning motion.
  • Higher inertia means more effort is required to change the spin rate.
Understanding this helps make sense of how easily or difficultly a spinning object will precess, which ties into the calculation of precession rate.
Precession Rate Calculation
Precession refers to the slow, conical motion of the rotation axis of a spinning object like a top. The precession rate \( \Omega \) signifies how quickly this motion happens, determined by the balance between torque and angular momentum. The formula employed is:
  • \[ \Omega = \frac{\tau}{I \cdot \omega} \]
Putting the values from the exercise:
  • \(\tau = 0.098 \text{ Nm}\)
  • \(I = 5.0 \times 10^{-4} \text{ kg}\cdot\text{m}^2\)
  • \(\omega = 60\pi \text{ rad/s}\)
  • \(\Omega = \frac{0.098}{5.0 \times 10^{-4} \cdot 60\pi} \approx 1.04 \text{ rad/s}\)
Thus, the precession rate is approximately 1.04 radians per second. This figure indicates how fast the rotational axis of the top moves in a circular path, which ties back to the torque and inertia involved.

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Most popular questions from this chapter

In a long jump, an athlete leaves the ground with an initial angular momentum that tends to rotate her body forward, threatening to ruin her landing. To counter this tendency, she rotates her outstretched arms to "take up" the angular momentum (Fig. 11-18). In \(0.700 \mathrm{~s}\), one arm sweeps through \(0.500\) rev and the other arm sweeps through \(1.000\) rev. Treat each arm as a thin rod of mass \(4.0 \mathrm{~kg}\) and length \(0.60 \mathrm{~m}\), rotating around one end. In the athlete's reference frame, what is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

A disk with a rotational inertia of \(7.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) rotates like a merry-go-round while undergoing a variable torque given by \(\tau=(5.00+2.00 t) \mathrm{N} \cdot \mathrm{m}\). At time \(t=1.00 \mathrm{~s}\), its angular momentum is \(5.00 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its angular momentum at \(t=3.00 \mathrm{~s}\) ?

A \(140 \mathrm{~kg}\) hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of \(0.150 \mathrm{~m} / \mathrm{s}\). How much work must be done on the hoop to stop it?

A horizontal vinyl record of mass \(0.10 \mathrm{~kg}\) and radius \(0.10 \mathrm{~m}\) rotates freely about a vertical axis through its center with an angular speed of \(4.7 \mathrm{rad} / \mathrm{s}\). The rotational inertia of the record about its axis of rotation is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2} .\) A wad of wet putty of mass \(0.020 \mathrm{~kg}\) drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately after the putty sticks to it?

A car travels at \(80 \mathrm{~km} / \mathrm{h}\) on a level road in the positive direction of an \(x\) axis. Each tire has a diameter of \(66 \mathrm{~cm}\). Relative to a woman riding in the car and in unit-vector notation, what are the velocity \(\vec{v}\) at the (a) center, (b) top, and (c) bottom of the tire and the magnitude \(a\) of the acceleration at the (d) center, (e) top, and (f) bottom of each tire? Relative to a hitchhiker sitting next to the road and in unit-vector notation, what are the velocity \(\vec{v}\) at the \((\mathrm{g})\) center, (h) top, and (i) bottom of the tire and the magnitude \(a\) of the acceleration at the (j) center, (k) top, and (1) bottom of each tire?
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