/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 A top spins at 30 rev/s about an... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(1.04 \text{ rad/s}\). (b) The precession is counterclockwise when viewed from above.

Step by step solution

01

Identify the known variables

We are given the following values from the problem statement: \( \omega_s = 30 \text{ rev/s} = 30 \times 2\pi \text{ rad/s}\), \(I = 5.0 \times 10^{-4} \text{ kg} \cdot \text{m}^2\), \(m = 0.50 \text{ kg}\), \(d = 4.0 \text{ cm} = 0.04 \text{ m}\), and \(\theta = 30^\circ\).
02

Calculate the angular speed

Convert the angular speed from revolutions per second to radians per second. Since one revolution is \(2\pi\) radians, \( \omega_s = 30 \times 2\pi \approx 188.5 \text{ rad/s}\).
03

Calculate the gravitational torque

The gravitational torque \(\tau\) about the pivot point is given by \( \tau = mgd \sin(\theta) \). Substituting the known values gives \( \tau = 0.50 \times 9.81 \times 0.04 \times \sin(30^\circ)\), which calculates to \( \tau \approx 0.0981 \text{ N} \cdot \text{m}\).
04

Calculate the precession rate

The precession rate \( \Omega \) is given by \( \Omega = \frac{\tau}{I \omega_s} \). Substituting the calculated torque and angular speed gives \( \Omega = \frac{0.0981}{5.0 \times 10^{-4} \times 188.5} \approx 1.04 \text{ rad/s}\).
05

Determine the direction of the precession

The gyroscopic precession direction is given by the right-hand rule. Since the top spins clockwise from an overhead view, use your right hand to curl your fingers in the direction of spin (clockwise when viewed from above) so your thumb points in the direction of angular momentum. The external torque acts horizontally to the left, causing precession to be counterclockwise when viewed from above.

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Key Concepts

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

Rotational Inertia
Rotational inertia, often known as the moment of inertia, is a measure of how difficult it is to change the rotation of an object. Think of it as the rotational equivalent of mass in linear motion. It tells us how much torque is needed to achieve a desired angular acceleration. The formula for rotational inertia is \[ I = rac{1}{2} m r^2 \] where \( m \) is the mass of the object and \( r \) is the radius from the axis of rotation to the point where the mass is concentrated.

In our original exercise, the rotational inertia \( I \) is given as \( 5.0 \times 10^{-4} \text{ kg} \cdot \text{m}^2 \). This means it takes more effort to start or stop the spinning top than if it had a lower rotational inertia. Understanding rotational inertia is crucial when studying objects in rotational motion, as it directly affects the object's tendency to continue spinning.
  • Increased rotational inertia means more stability in the spinning motion.
  • Objects with larger distribution of mass from the axis of rotation have higher rotational inertia.
Angular Momentum
Angular momentum is a concept that describes the tendency of a rotating object to continue spinning unless acted upon by an external torque. It is the product of the rotational inertia and the angular velocity, denoted mathematically as \( L = I \omega \) where \( L \) is the angular momentum, \( I \) is the rotational inertia, and \( \omega \) is the angular velocity of the object. Just like linear momentum, angular momentum is a vector quantity, which means it has both a magnitude and a direction. The direction of angular momentum is determined by the right-hand rule: if the fingers of your right hand curl in the direction of rotation, your thumb points in the direction of the angular momentum vector. In the given exercise, the top spins at an angular speed \( \omega = 188.5 \text{ rad/s} \), indicating a high angular momentum which resists changes in its spinning state.

Key features of angular momentum include:
  • It's conserved in a closed system with no external torques.
  • Higher angular momentum means greater stability against external influences.
Gravitational Torque
Gravitational torque is the measure of the rotational force exerted by gravity around a pivot point or axis. In simpler terms, it is the twisting force caused by the weight of an object acting at a distance from the pivot. The formula for calculating gravitational torque \( \tau \) is \[ \tau = mgd \sin(\theta) \] where \( m \) is the mass, \( g \) is the gravitational acceleration, \( d \) is the distance from the pivot point, and \( \theta \) is the angle with the vertical.

In the provided exercise, the torque calculated is approximately \( 0.0981 \text{ N} \cdot \text{m} \). This torque is responsible for the precession observed in the spinning top. Gravitational torque plays a crucial role in gyroscopic effects, affecting the direction and rate of precession.
  • Torque is higher when the mass is further from the pivot point.
  • The angle \( \theta \) affects the sine component, influencing the overall magnitude of torque.
  • Torque can cause precession, changing how objects maintain rotational direction.

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Most popular questions from this chapter

A Cylindrical object of mass \(M\) and radius \(R\) rolls smoothly from rest down a ramp and onto a horizontal section. From there it rolls off the ramp and onto the floor, landing a horizontal distance \(d=0.506 \mathrm{~m}\) from the end of the ramp. The initial height of the object is \(H=0.90 \mathrm{~m}\) the end of the ramp is at height \(h=0.10 \mathrm{~m}\). The object consists of an outer cylindrical shell (of a certain uniform density) that is glued to a central cylinder (of a different uniform density). The rotational inertia of the object can be expressed in the general form \(I=\beta M R^{2},\) but \(\beta\) is not 0.5 as it is for a cylinder of uniform density. Determine \(\beta\)

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