Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 68

A 0.050 -kg bead slides on a wire bent into a circle of radius \(0.40 \mathrm{~m}\). You pluck the bead with a force tangent to the circle. What force is needed to give the bead an angular acceleration of \(6.0 \mathrm{rad} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
Answer: The force needed to achieve the given angular acceleration is 0.12 N.

Step by step solution

01

Write down Newton's second law for rotational motion

Newton's second law for rotational motion is given by: \(\tau = I \alpha\) Where \(\tau\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration.
02

Find the moment of inertia of the bead

Since the bead is moving on a circular path with a constant radius, we can treat it as a point mass and use the formula for the moment of inertia of a point mass: \(I = mr^2\) where \(m\) is the mass of the bead, and \(r\) is the radius of the circle. Plugging in the given values, we get: \(I = (0.050\,\text{kg})(0.40\,\text{m})^2 = 0.008\,\text{kg}\cdot\text{m}^2\)
03

Calculate the torque needed to achieve the given angular acceleration

Now that we have the moment of inertia, we can use Newton's second law for rotational motion to find the torque needed to achieve the given angular acceleration: \(\tau = I \alpha\) Plugging in the values for \(I\) and \(\alpha\), we get: \(\tau = (0.008\,\text{kg}\cdot\text{m}^2)(6.0\,\text{rad}/\text{s}^2) = 0.048\,\text{N}\cdot\text{m}\)
04

Determine the tangential force needed to produce the required torque

Since the force is applied tangentially to the bead, the torque produced by the force can be calculated as: \(\tau = r F_\text{t}\) where \(F_\text{t}\) is the tangential force. We can now solve for \(F_\text{t}\): \(F_\text{t} = \frac{\tau}{r}\) Substituting the values, we get: \(F_\text{t} = \frac{0.048\,\text{N}\cdot\text{m}}{0.40\,\text{m}} = 0.12\,\text{N}\) Therefore, the force needed to give the bead an angular acceleration of \(6.0\,\text{rad}/\text{s}^2\) is \(0.12\,\text{N}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A ballistic pendulum consists of an arm of mass \(M\) and length \(L=0.48 \mathrm{~m} .\) One end of the arm is pivoted so that the arm rotates freely in a vertical plane. Initially, the arm is motionless and hangs vertically from the pivot point. A projectile of the same mass \(M\) hits the lower end of the arm with a horizontal velocity of \(V=3.6 \mathrm{~m} / \mathrm{s}\). The projectile remains stuck to the free end of the arm during their subsequent motion. Find the maximum angle to which the arm and attached mass will swing in each case: a) The arm is treated as an ideal pendulum, with all of its mass concentrated as a point mass at the free end. b) The arm is treated as a thin rigid rod, with its mass evenly distributed along its length.

A child builds a simple cart consisting of a \(60.0 \mathrm{~cm}\) by \(1.20 \mathrm{~m}\) sheet of plywood of mass \(8.00 \mathrm{~kg}\) and four wheels, each \(20.0 \mathrm{~cm}\) in diameter and with a mass of \(2.00 \mathrm{~kg}\). It is released from the top of a \(15.0^{\circ}\) incline that is \(30.0 \mathrm{~m}\) long. Find the speed at the bottom. Assume that the wheels roll along the incline without slipping and that friction between the wheels and their axles can be neglected.

A uniform solid sphere of radius \(R, \operatorname{mass} M,\) and moment of inertia \(I=\frac{2}{5} M R^{2}\) is rolling without slipping along a horizontal surface. Its total kinetic energy is the sum of the energies associated with translation of the center of mass and rotation about the center of mass. Find the fraction of the sphere's total kinetic energy that is attributable to rotation.

Many pulsars radiate radio frequency or other radiation in a periodic manner and are bound to a companion star in what is known as a binary pulsar system. In \(2004,\) a double pulsar system, PSR J0737-3039A and J0737-3039B, was discovered by astronomers at the Jodrell Bank Observatory in the United Kingdom. In this system, both stars are pulsars. The pulsar with the faster rotation period rotates once every \(0.023 \mathrm{~s}\), while the other pulsar has a rotation period of \(2.8 \mathrm{~s}\). The faster pulsar has a mass 1.337 times that of the Sun, while the slower pulsar has a mass 1.250 times that of the Sun. a) If each pulsar has a radius of \(20.0 \mathrm{~km}\), express the ratio of their rotational kinetic energies. Consider each star to be a uniform sphere with a fixed rotation period. b) The orbits of the two pulsars about their common center of mass are rather eccentric (highly squashed ellipses), but an estimate of their average translational kinetic energy can be obtained by treating each orbit as circular with a radius equal to the mean distance from the system's center of mass. This radius is equal to \(4.23 \cdot 10^{8} \mathrm{~m}\) for the larger star, and \(4.54 \cdot 10^{8} \mathrm{~m}\) for the smaller star. If the orbital period is \(2.4 \mathrm{~h},\) calculate the ratio of rotational to translational kinetic energies for each star.

A solid sphere rolls without slipping down an incline, starting from rest. At the same time, a box starts from rest at the same altitude and slides down the same incline, with negligible friction. Which object arrives at the bottom first? a) The solid sphere arrives first. b) The box arrives first. c) Both arrive at the same time. d) It is impossible to determine.
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Work Energy and Power

Read Explanation

Turning Points in Physics

Read Explanation

Energy Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electricity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free