/*! 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 84 Two uniform solid spheres, each ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two uniform solid spheres, each with mass \(M=0.800 \mathrm{~kg}\) and radius \(R=0.0800 \mathrm{~m},\) are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant \(k=160 \mathrm{~N} / \mathrm{m}\) has one end attached to the wall and the other end attached to a friction less ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the center of mass of the spheres is simple harmonic and calculate the period.

Short Answer

Expert verified
The motion of the center of mass of the spheres can be shown to be a simple harmonic motion with a period given by \(T = 2\pi \sqrt{ \frac{m + \frac{2}{5}m}{k}}\).

Step by step solution

01

Identify the Elements of the Problem

From the problem, we identify that the mass of each of the spheres is \(M=0.800 \mathrm{~kg}\), the radius is \(R=0.0800 \mathrm{~m}\), and the spring constant is \(k=160 \mathrm{~N} / \mathrm{m}\). We also know that the moment of inertia of a sphere about any diameter is given by \(I = \frac{2}{5}MR^2\). The spheres are attached at their centers by a rod. Therefore, the total moment of inertia of this system will be \(2 \cdot I = \frac{2}{5}2MR^2 = \frac{4}{5}MR^2\). The problem states that the spheres roll without slipping.
02

Apply the Conservation Principle

Given that the spheres roll without slipping, this implies that the total energy of the system is constant and sum of the potential and kinetic energy is also conserved. The potential energy of the spring is given by \(\frac{1}{2}kA^2\), where \(A\) is the amplitude (maximum stretch or compression) of the spring. When the spheres are at rest (i.e., have zero kinetic energy), the spring is maximally stretched or compressed. The kinetic energy of the spheres when the spring has zero potential energy is given by \(\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\), where \(\omega = v/R\) is the angular velocity of the spheres.
03

Calculate the Total Energy of the System

Equating the kinetic and potential energy gives \(\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{4}{5}MR^2)(v/R)^2\). Simplifying this relation, we get \(v = \sqrt{ \frac{kA^2}{m + \frac{2}{5}m}}\).
04

Calculate the Time Period of the Simple Harmonic Motion

The speed at \(x = 0\) in a simple harmonic motion is given by \(v = \omega A\), where \(\omega = 2\pi / T\) is the angular frequency and \(T\) is the period of the motion. Equating the expressions for \(v\) gives \(\sqrt{ \frac{kA^2}{m + \frac{2}{5}m}} = \frac{2\pi A}{T}\). Solving for \(T\) gives \(T = 2\pi \sqrt{ \frac{m + \frac{2}{5}m}{k}}\).

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 91Ó°ÊÓ!

Key Concepts

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

Mechanics of Rolling Objects
Understanding the mechanics of rolling objects like the solid spheres in our exercise is fundamental to physics. When an object rolls without slipping, a static friction force is at play. This force prevents the surface points of the object in contact with the ground from sliding.

As the objects roll, each point on the sphere in contact with the surface momentarily has zero velocity relative to the surface, and the sphere's motion combines translation (straight-line movement) of its center of mass and rotation around its center of mass.

To describe such motion, we define 'rolling without slipping' as a condition where for any point of time, the linear velocity at the center of mass, denoted by \(v\), is equal to the radius \(R\) times the angular velocity \(\omega\), which is articulated through the equation \(v = R\omega\). This creates a direct link between translational and rotational motion, and analyzing one can provide information about the other.
Moment of Inertia
The moment of inertia, symbolized by \(I\), is a quantity expressing an object's tendency to resist angular acceleration. It is governed by the distribution of mass within the object and the axis about which it rotates. For a solid sphere, the moment of inertia about any diameter is \(I = \frac{2}{5}MR^2\), where \(M\) is the mass and \(R\) is the radius of the sphere.

In our problem's context, we have two identical solid spheres, so each contributes equally to the system's overall moment of inertia. Since they're connected by a light rod, which we assume has a negligible moment of inertia, the combined moment of inertia for the system essentially doubles that of a single sphere, resulting in \(I_{total} = \frac{4}{5}MR^2\).

When the spheres roll without slipping, their rotational motion about the rod contributes to the system's energy dynamics, critically affecting the behavior during spring oscillation. The moment of inertia is pivotal in calculating the kinetic energy associated with the rotating mass.
Conservation of Energy
The principle of conservation of energy states that within a closed system, energy cannot be created or destroyed but can only be transformed from one form to another. In the scenario of the rolling spheres, the system's total energy fluctuates between kinetic energy and potential energy of the spring.

When the spring is stretched or compressed, it stores potential energy given by \(\frac{1}{2}kA^2\). As the spring returns to its equilibrium position, this potential energy converts to kinetic energy. The kinetic energy comprises translational kinetic energy \(\frac{1}{2}mv^2\) and rotational kinetic energy \(\frac{1}{2}I\omega^2\). Because the spheres roll without slipping, we know their kinetic energies are always linked through the relationship \(\omega = \frac{v}{R}\).

By equating the kinetic energy with the potential energy at the maximal stretch or compression, we can solve for variables such as velocity at equilibrium, and eventually find the period of simple harmonic motion using \(T = 2\pi \sqrt{ \frac{m + \frac{2}{5}m}{k}}\). This overall energy conservation is a core tenet in analyzing the motion of any system and is critical for understanding the behavior of the rolling spheres attached to the spring in our exercise.

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 block of mass \(m\) is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring. The spring has force constant \(k\), and the amplitude of the \(\mathrm{SHM}\) is \(A\). The block has \(v=0,\) and \(x=+A\) at \(t=0 .\) It first reaches \(x=0\) when \(t=T / 4\) where \(T\) is the period of the motion. (a) In terms of \(T,\) what is the time \(t\) when the block first reaches \(x=A / 2 ?\) (b) The block has its maximum speed when \(t=T / 4\). What is the value of \(t\) when the speed of the block first reaches the value \(v_{\max } / 2 ?\) (c) Does \(v=v_{\max } / 2\) when \(x=A / 2 ?\)

SHM of a Floating Object. An object with height \(h\)mass \(M\), and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\). (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M,\) and the cross-sectional area \(A\) of the object. You can ignore the damping due to fluid friction (see Section 14.7).

A holiday ornament in the shape of a hollow sphere with mass \(M=0.015 \mathrm{~kg}\) and radius \(R=0.050 \mathrm{~m}\) is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (Hint: Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

In a physics lab, you attach a \(0.200 \mathrm{~kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is \(2.60 \mathrm{~s}\). Find the spring's force constant.

A thin metal disk with mass \(2.00 \times 10^{-3} \mathrm{~kg}\) and radius \(2.20 \mathrm{~cm}\) is attached at its center to a long fiber (Fig. \(\mathbf{E 1 4 . 4 0}\) ). The disk, when twisted and released, oscillates with a period of \(1.00 \mathrm{~s}\). Find the torsion constant of the fiber.

See all solutions

Recommended explanations on Physics Textbooks

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.