/*! 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 19 In another race, a solid sphere ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 19

In another race, a solid sphere and a thin ring roll without slipping from rest down a ramp that makes angle \(\theta\) with the horizontal. Find the ratio of their accelerations, \(a_{\text {ring }} / a_{\text {sphere }}\)

Short Answer

Expert verified
Answer: The ratio of accelerations of a solid sphere and a thin ring rolling without slipping down a ramp at angle 胃 with the horizontal is 7/10.

Step by step solution

01

1. Write down the expressions of the forces acting on the objects

The gravitational force \(F_g\) acts vertically downwards on both the sphere and the ring. It can be decomposed into two components; one along the incline (\(F_\parallel\) which causes the object to roll) and one perpendicular to the incline (\(F_\perp\) which doesn't contribute to the motion). We will find \(F_\parallel\) for both objects. For both objects, we have: \(F_g = mg\) where \(m\) is the mass of the object (sphere or ring) and \(g\) is the acceleration due to gravity. The component of gravitational force along the incline, in terms of \(\theta\) is given by: \(F_\parallel = mg\sin\theta\)
02

2. Set up equations of motion

For rolling without slipping, we need to equate the translational and rotational motion. We use Newton's second law for translational motion and the torque equation for rotational motion: For translational motion, we have: $$F_\parallel = ma$$ For rotational motion, we have: $$I\alpha=R \cdot F_\parallel$$ where \(I\) is the moment of inertia of the object, \(\alpha\) is the angular acceleration, and \(R\) is the radius. Since the objects are rolling without slipping, we can relate the linear and angular acceleration using the formula: $$a=R\alpha$$
03

3. Calculate the moment of inertia for each object

Now, we need to find the moment of inertia, \(I\), for each object. The moment of inertia depends on the distribution of mass in the object. For a solid sphere of mass \(m\) and radius \(R\), its moment of inertia about an axis through the center is given by: \(I_{\text {sphere }}=\frac{2}{5}mR^2\) For a thin ring of mass \(m\) and radius \(R\), its moment of inertia about an axis perpendicular to the plane of the ring and passing through the center is given by: \(I_{\text {ring }}=mR^2\)
04

4. Solve the equations for the accelerometer and find the ratio

Now we need to solve the equations of motion for both objects to find the accelerations for each object: For the sphere: $$\begin{aligned} m a_{\text {sphere }} &=m g \sin \theta \\ \frac{2}{5} m R^2 \alpha_{\text {sphere }} &=R \cdot a_{\text {sphere }} \\ a_{\text{sphere}} &= \frac{5}{7}g\sin\theta \end{aligned}$$ For the ring: $$\begin{aligned} m a_{\text {ring }} &=m g \sin \theta \\ m R^2 \alpha_{\text {ring }} &=R \cdot a_{\text {ring }} \\ a_{\text{ring}} &= \frac{1}{2}g\sin\theta \end{aligned}$$ Now we can find the desired ratio: $$\frac{a_{\text {ring }} }{a_{\text {sphere }}}=\frac{\frac{1}{2} g \sin \theta}{\frac{5}{7} g \sin \theta}=\boxed{\frac{7}{10}}$$

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

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 thin uniform rod (length \(=1.00 \mathrm{~m},\) mass \(=2.00 \mathrm{~kg})\) is pivoted about a horizontal frictionless pin through one of its ends. The moment of inertia of the rod through this axis is \(\frac{1}{3} m L^{2} .\) The rod is released when it is \(60.0^{\circ}\) below the horizontal. What is the angular acceleration of the rod at the instant it is released?

An ice skater rotating on frictionless ice brings her hands into her body so that she rotates faster. Which, if any, of the conservation laws hold? a) conservation of mechanical energy and conservation of angular momentum b) conservation of mechanical energy only c) conservation of angular momentum only d) neither conservation of mechanical energy nor conservation of angular momentum

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 \(6.0 \mathrm{rad} / \mathrm{s}^{2} ?\)

A 25.0 -kg boy stands \(2.00 \mathrm{~m}\) from the center of a frictionless playground merry-go-round, which has a moment of inertia of \(200 . \mathrm{kg} \mathrm{m}^{2} .\) The boy begins to run in a circular path with a speed of \(0.600 \mathrm{~m} / \mathrm{s}\) relative to the ground. a) Calculate the angular velocity of the merry-go-round. b) Calculate the speed of the boy relative to the surface of the merry-go- round.
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Magnetism

Read Explanation

Classical Mechanics

Read Explanation

Rotational Dynamics

Read Explanation

Electromagnetism

Read Explanation

Electrostatics

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.