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Chapter 10: Problem 66

A uniform spherical shell of mass \(M=4.5 \mathrm{~kg}\) and radius \(R=8.5 \mathrm{~cm}\) can rotate about a vertical axis on frictionless bearings (Fig. 10-44). A massless cord passes around the equator of the shell, over a pulley of rotational inertia \(I=3.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and radius \(r=5.0 \mathrm{~cm}\), and is attached to a small object of mass \(m=0.60 \mathrm{~kg}\). There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen \(82 \mathrm{~cm}\) after being released from rest? Use energy considerations.

Short Answer

Expert verified
The speed of the object when it has fallen is approximately 2.3 m/s.

Step by step solution

01

Identify Energy Types

We need to consider the kinetic energy of the system and the gravitational potential energy. The kinetic energy includes the translational kinetic energy of the descending mass and the rotational kinetic energies of the spherical shell and the pulley. Initially, the potential energy is at a maximum when the mass is held at height.
02

Write Energy Conservation Equation

The principle of conservation of energy states that the total energy remains constant. Thus, \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I_{shell}\omega^2 + \frac{1}{2}I_{pulley}\Omega^2 \] where:- \( I_{shell} = \frac{2}{3} MR^2 \), the moment of inertia of the spherical shell- \( \omega \) and \( \Omega \) are angular speeds for shell and pulley. Since the cord does not slip, \( v = R\omega = r\Omega \).
03

Substitute Known Values and Expressions

Substitute \( \omega = \frac{v}{R} \) and \( \Omega = \frac{v}{r} \) into the equation. The expression becomes:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2}\left(\frac{2}{3} MR^2\right)\left(\frac{v}{R}\right)^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2 \]
04

Simplify the Equation

Simplifying the equation, we have:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{3} Mv^2 + \frac{1}{2}\left(\frac{I}{r^2} \right)v^2 \]Combine all the terms on the right:\[ mgh = \left(\frac{1}{2} m + \frac{1}{3} M + \frac{1}{2}\frac{I}{r^2} \right)v^2 \]
05

Solve for Speed \(v\)

Rearrange the equation to solve for \(v\):\[ v = \sqrt{\frac{2mgh}{m + \frac{2}{3}M + \frac{I}{r^2}}} \]Substitute \( m = 0.60 \text{ kg} \), \( h = 0.82 \text{ m} \), \( M = 4.5 \text{ kg} \), \( R = 0.085 \text{ m} \), \( r = 0.050 \text{ m} \), and \( I = 3.0 \times 10^{-3} \mathrm{~kg} \, m^2 \).
06

Calculate Final Speed

Plug in the values:\[ v = \sqrt{\frac{2 \times 0.60 \times 9.8 \times 0.82}{0.60 + \frac{2}{3}\times 4.5 + \frac{3.0 \times 10^{-3}}{(0.050)^2}}} \]Perform the calculations:\[ v \approx 2.3 \text{ m/s} \]

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Key Concepts

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

Conservation of Energy
Conservation of energy is a fundamental principle in physics stating that energy cannot be created or destroyed, only transformed. In this exercise, we analyze the energy transformations that occur when an object falls.

Initially, the system has gravitational potential energy, given by the equation \( mgh \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( h \) is the height. As the object falls, this potential energy is converted into kinetic energy 鈥� both translational and rotational kinetic energies. The total mechanical energy at the beginning and at the end of the motion remains constant.
  • Gravitational potential energy transforms into kinetic energy.
  • Total energy remains constant: \( E_{initial} = E_{final} \).
The conservation of energy equation helps us equate these energies to solve for the final speed of the object.
Moment of Inertia
Moment of inertia is a property of physical objects that quantifies their resistance to rotational acceleration. It's similar to mass in linear motion, but instead of affecting inertia in straight lines, it affects rotation.

For the spherical shell in this problem, the moment of inertia \( I_{shell} \) is given by \( \frac{2}{3}MR^2 \), where \( M \) is the mass and \( R \) is the radius.
  • Higher moment of inertia means the object is harder to rotate.
  • Dependence on object's shape and mass distribution.
This parameter is crucial when calculating rotational kinetic energies in the energy conservation equation.
Translational Kinetic Energy
Translational kinetic energy refers to the energy due to the motion of the center of mass of an object. This type of kinetic energy is calculated using the formula \( \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity.

In this exercise, the mass attached to the cord experiences translational motion as it falls and contributes this form of kinetic energy to the system. Its translational energy is part of the total kinetic energy considered in the energy conservation equation.
  • Directly proportional to the mass and the square of the velocity.
  • A vital component when analyzing linear motion energy.
Angular Velocity
Angular velocity describes how fast an object is rotating. It is the rate at which an object rotates or revolves relative to another point, usually the center of a circular path.

In this system, the angular velocities \( \omega \) for the shell and \( \Omega \) for the pulley are important in linking linear and rotational motion. Using \( v = R\omega = r\Omega \) provides a way to connect translational speed \( v \) to the angular velocities.
  • Measured in radians per second.
  • Key in converting between linear and rotational motions.
Angular velocity is essential in ensuring that all forms of kinetic energy are correctly accounted for in the energy equations.

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Most popular questions from this chapter

Four identical particles of mass \(0.50 \mathrm{~kg}\) each are placed at the vertices of a \(2.0 \mathrm{~m} \times 2.0 \mathrm{~m}\) square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square, (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?

The angular position of a point on the rim of a rotating wheel is given by \(\theta=4.0 t-3.0 t^{2}+t^{3}\), where \(\theta\) is in radians and \(t\) is in seconds. What are the angular velocities at (a) \(t=2.0 \mathrm{~s}\) and \((\mathrm{b}) t=4.0 \mathrm{~s} ?(\mathrm{c})\) What is the average angular acceleration for the time interval that begins at \(t=2.0 \mathrm{~s}\) and ends at \(t=4.0 \mathrm{~s}\) ? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?

Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of \(200 \pi \mathrm{rad} / \mathrm{s}\). One such flywheel is a solid, uniform cylinder with a mass of \(500 \mathrm{~kg}\) and a radius of \(1.0 \mathrm{~m} .\) (a) What is the kinetic energy of the flywheel after charging? (b) If the truck uses an average power of \(8.0 \mathrm{~kW}\), for how many minutes can it operate between chargings?

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at \(10 \mathrm{rev} / \mathrm{s} ; 60\) revolutions later, its angular speed is \(15 \mathrm{rev} / \mathrm{s}\). Calculate (a) the angular acceleration, (b) the time required to complete the 60 revolutions, (c) the time required to reach the 10 rev/s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 10 rev/s angular speed.

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