Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 2

Two solid steel balls, one small and one large, are on an inclined plane. The large ball has a diameter twice as large as that of the small ball. Starting from rest, the two balls roll without slipping down the incline until their centers of mass are \(1 \mathrm{~m}\) below their starting positions. What is the speed of the large ball \(\left(v_{\mathrm{L}}\right)\) relative to that of the small ball \(\left(v_{\mathrm{S}}\right)\) after rolling \(1 \mathrm{~m} ?\) a) \(v_{\mathrm{L}}=4 v_{\mathrm{S}}\) d) \(v_{\mathrm{L}}=0.5 v_{\mathrm{S}}\) b) \(v_{\mathrm{L}}=2 v_{\mathrm{S}}\) e) \(v_{\mathrm{L}}=0.25 v_{\mathrm{S}}\) c) \(v_{\mathrm{L}}=v_{\mathrm{S}}\)

Short Answer

Expert verified
a) v_L = v_S b) v_L = 2v_S c) v_L = 4v_S d) v_L = 0.5 v_S

Step by step solution

01

1. Find the moment of inertia of the solid sphere

For a solid sphere, the moment of inertia (I) is given by the formula: \(I = \frac{2}{5}mr^2\), where m is the mass and r is the radius of the sphere.
02

2. Set up an equation for conservation of mechanical energy

As the balls roll without slipping, conservation of mechanical energy states that the total mechanical energy at the beginning (potential energy) equals the total mechanical energy at the end (kinetic energy + rotational energy). Thus, we have: \(mgh = \frac{1}{2}mv^2 + \frac{1}{2}I{\omega}^2\) , where m: mass of the ball, g: acceleration due to gravity, h: height of the ball, v: linear velocity of the ball, I: moment of inertia, ω: angular velocity of the ball.
03

3. Relate linear velocity and angular velocity

The relationship between linear velocity (v) and angular velocity (ω) for a rolling sphere without slipping is given by: \(v = r\omega\). This relationship will be used to eliminate ω from the conservation of mechanical energy equation.
04

4. Apply the conservation of mechanical energy equation to both balls and divide the equations

For the small ball, we have: \(mg(1) = \frac{1}{2}m(v_{S})^2 + \frac{1}{2}(\frac{2}{5}mr^2)({\frac{v_{S}}{r}})^2\) And for the large ball, we have: \(mg(1) = \frac{1}{2}m(v_{L})^2 + \frac{1}{2}(\frac{2}{5}m(2r)^2)({\frac{v_{L}}{2r}})^2\) Now, we divide the equation for the large ball by the equation for the small ball to find the ratio of their velocities, \((\)v_L/v_S\()\):
05

5. Solve for the ratio of the velocities

After dividing the equations and simplifying, we get: \(\frac{v_{L}^2}{v_{S}^2} = \frac{1}{2}\) Taking the square root of both sides, we have: \(v_{L} = \sqrt{\frac{1}{2}}v_{S}\) Comparing with the given options, we have \(v_{\mathrm{L}} = 0.5 v_{\mathrm{S}}\). Therefore, the correct answer is (d) \(v_{\mathrm{L}} = 0.5 v_{\mathrm{S}}\).

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 wagon wheel is made entirely of wood. Its components consist of a rim, 12 spokes, and a hub. The rim has mass \(5.2 \mathrm{~kg}\), outer radius \(0.90 \mathrm{~m}\), and inner radius \(0.86 \mathrm{~m}\). The hub is a solid cylinder with mass \(3.4 \mathrm{~kg}\) and radius \(0.12 \mathrm{~m} .\) The spokes are thin rods of mass \(1.1 \mathrm{~kg}\) that extend from the hub to the inner side of the rim. Determine the constant \(c=I / M R^{2}\) for this wagon wheel.

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.

A uniform solid sphere of mass \(m\) and radius \(r\) is placed on a ramp inclined at an angle \(\theta\) to the horizontal. The coefficient of static friction between sphere and ramp is \(\mu_{s} .\) Find the maximum value of \(\theta\) for which the sphere will roll without slipping, starting from rest, in terms of the other quantities.

A circular platform of radius \(R_{p}=4.00 \mathrm{~m}\) and mass \(M_{\mathrm{p}}=400 .\) kg rotates on frictionless air bearings about its vertical axis at 6.00 rpm. An 80.0 -kg man standing at the very center of the platform starts walking \((\) at \(t=0)\) radially outward at a speed of \(0.500 \mathrm{~m} / \mathrm{s}\) with respect to the platform. Approximating the man by a vertical cylinder of radius \(R_{\mathrm{m}}=0.200 \mathrm{~m}\) determine an equation (specific expression) for the angular velocity of the platform as a function of time. What is the angular velocity when the man reaches the edge of the platform?

Why does a figure skater pull in her arms while increasing her angular velocity in a tight spin?
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Famous Physicists

Read Explanation

Geometrical and Physical Optics

Read Explanation

Physical Quantities And Units

Read Explanation

Wave Optics

Read Explanation

Linear Momentum

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