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Chapter 11: Problem 68

A top spins at 30 rev/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) The precession rate is approximately 1.041 rad/s. (b) The direction of precession is counterclockwise from an overhead view.

Step by step solution

01

Calculate the Torque due to Gravity

The torque due to gravity, \( \tau \), is given by the formula \( \tau = m \cdot g \cdot d \cdot \sin(\theta) \), where \( m = 0.50 \) kg is the mass, \( g = 9.8 \) m/s虏 is the gravitational acceleration, \( d = 0.04 \) m is the distance from the pivot, and \( \theta = 30^{\circ} \) is the angle. Therefore, \( \tau = 0.50 \cdot 9.8 \cdot 0.04 \cdot \sin(30^{\circ}) = 0.098 \) Nm.
02

Determine the Angular Momentum

The angular momentum \( L \) of the top due to its spin is \( L = I \cdot \omega_s \), where \( I = 5.0 \times 10^{-4} \) kg路m虏 is the rotational inertia and \( \omega_s = 2\pi \cdot 30 \) rad/s is the angular velocity. Thus, \( L = 5.0 \times 10^{-4} \cdot 60\pi \approx 0.0942 \) kg路m虏/s.
03

Calculate the Precession Rate

The precession rate \( \Omega \) is given by \( \Omega = \frac{\tau}{L} \). Substituting values, we have \( \Omega = \frac{0.098}{0.0942} \approx 1.041 \) rad/s.
04

Determine the Direction of Precession

The direction of precession is determined by the right-hand rule. Since the top spins clockwise, the angular momentum vector \( L \) points downward. The torque causes \( L \) to move horizontally in a direction given by the right-hand rule, causing a counterclockwise precession when viewed from above.

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Key Concepts

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

Torque
Torque is a measure of the rotational force applied on an object, which causes it to rotate around an axis. It can be thought of as a twisting or turning force. The amount of torque depends primarily on three factors: the magnitude of the force applied, the distance from the pivot point, and the angle at which the force is applied.
For the formula, torque is represented as \( \tau = m \cdot g \cdot d \cdot \sin(\theta) \), where
  • \( \tau \) is the torque,
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity (\( 9.8 \) m/s虏),
  • \( d \) is the lever arm distance from the pivot point,
  • \( \theta \) is the angle between the force and the lever arm direction.
Torque plays a crucial role in determining how an object starts to spin or change its spinning speed. In precession, like the spinning top, torque from gravity helps cause the top to precess or wobble around its axis. Without torque, the top would just spin around its central axis without any wobbling motion.
Angular Momentum
Angular momentum is the rotational equivalent of linear momentum and is a vector quantity, which signifies how much rotation an object has and in which direction. It is a conserved quantity in a closed system, meaning the total angular momentum remains constant if there are no external torques.
The general formula for angular momentum \( L \) is\( L = I \cdot \omega \), where
  • \( L \) is the angular momentum,
  • \( I \) is the rotational inertia of the object,
  • \( \omega \) is the angular velocity (measured in radians per second).
In the example of a spinning top, the angular momentum helps determine the direction and stability of the spin. This concept is key to understanding why the top doesn't simply fall over but instead moves around in a way that looks like it's wobbling or precessing. The spin direction is related to the angular momentum vector by a rule called the right-hand rule, which helps in analyzing rotational directions and effects.
Rotational Inertia
Rotational inertia, also known as the moment of inertia, is a property that measures how much resistance an object has against changes to its rotational motion. It's like mass in linear motion but for rotating bodies.
In mathematics, it is expressed using the formula\( I = \sum{m_i \cdot r_i^2} \), where
  • \( I \) is the rotational inertia,
  • \( m_i \) is the mass of each particle within the object,
  • \( r_i \) is the distance of each particle from the axis of rotation.
A higher rotational inertia means the object is harder to start or stop rotating. For the spinning top, the specified rotational inertia around its central axis determines how easily it spins swiftly without wobbling.
This factor is essential when discussing how rotating objects behave under various forces and moments such as those causing them to precess. It plays a vital role when calculating other rotational dynamics, such as angular momentum, particularly when an axis is involved in complex spinning motions.

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Most popular questions from this chapter

During a jump to his partner, an aerialist is to make a quadruple somersault lasting a time \(t=1.87 \mathrm{~s}\). For the first and last quarter- revolution, he is in the extended orientation shown in Fig. \(11-55\), with rotational inertia \(I_{1}=19.9 \mathrm{~kg} \cdot \mathrm{m}^{2}\) around his center of mass (the dot). During the rest of the flight he is in a tight tuck, with rotational inertia \(I_{2}=3.93 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) What must be his angular speed \(\omega_{2}\) around his center of mass during the tuck?

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, \((\mathrm{k})\) top, and (1) bottom of each tire?

At time \(t=0\), a \(3.0 \mathrm{~kg}\) particle with velocity \(\vec{v}=(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(6.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) is at \(x=3.0 \mathrm{~m}, y=8.0 \mathrm{~m} .\) It is pulled by a \(7.0 \mathrm{~N}\) force in the negative \(x\) direction. About the origin, what are (a) the particle's angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?

Figure 11-51 shows an overhead view of a ring that can rotate about its center like a merrygo-round. Its outer radius \(R_{2}\) is \(0.800 \mathrm{~m}\), its inner radius \(R_{1}\) is \(R_{2} / 2.00\), its mass \(M\) is \(8.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2}\). By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1}\) ?

A bowler throws a bowling ball of radius \(R=11 \mathrm{~cm}\) along a lane. The ball (Fig. \(11-38\) ) slides on the lane with initial speed \(v_{\mathrm{com}, 0}=8.5 \mathrm{~m} / \mathrm{s}\) and initial angular speed \(\omega_{0}=0 .\) The coefficient of kinetic friction between the ball and the lane is \(0.21\). The kinetic frictional force \(\vec{f}_{k}\) acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed \(v_{\text {com }}\) has decreased enough and angular speed \(\omega\) has increased enough, the ball stops sliding and then rolls smoothly. (a) What then is \(v_{\text {com }}\) in terms of \(\omega\) ? During the sliding, what are the ball's (b) linear acceleration and (c) angular acceleration? (d) How long does the ball slide? (e) How far does the ball slide? (f) What is the linear speed of the ball when smooth rolling begins?
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