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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. 10-38). 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 is approximately 1.43 m/s.

Step by step solution

01

Define the System

We have a uniform spherical shell of mass \(M = 4.5 \mathrm{\,kg}\) and radius \(R = 8.5 \mathrm{\,cm}\), and a small object of mass \(m = 0.60 \mathrm{\,kg}\), connected by a cord over a pulley with rotational inertia \(I = 3.0 \times 10^{-3} \mathrm{\,kg \cdot m^2}\) and radius \(r = 5.0 \mathrm{\,cm}\). The object falls a distance \(h = 82 \mathrm{\,cm}\).
02

Apply Conservation of Energy

When the small object falls, the potential energy it loses is converted into kinetic energy of the object, rotational kinetic energy of the sphere, and rotational kinetic energy of the pulley. Initial total energy is given by the gravitational potential energy: \[PE_i = mgh = 0.60 \times 9.8 \times 0.82 \; \text{J}.\]
03

Express Final Energy in the System

The final mechanical energy consists of the translational kinetic energy of the object, the rotational kinetic energy of the sphere, and the rotational kinetic energy of the pulley: \[KE_f = \frac{1}{2}mv^2 + \frac{1}{2}I_{shell}\omega_{shell}^2 + \frac{1}{2}I_{pulley}\omega_{pulley}^2.\]
04

Relate Velocities and Angular Velocities

Since the cord does not slip, the velocity \(v\) of the small object is related to the angular velocity \(\omega_{shell}\) of the sphere and \(\omega_{pulley}\) of the pulley by \(v = R\omega_{shell} = r\omega_{pulley}\). So, \(\omega_{shell} = \frac{v}{R}\) and \(\omega_{pulley} = \frac{v}{r}\).
05

Use the Moment of Inertia of the Shell

The moment of inertia of a uniform spherical shell is \(I_{shell} = \frac{2}{3}MR^2\). Substitute this into the expression for rotational kinetic energy:\[KE_{shell} = \frac{1}{2}\left(\frac{2}{3}MR^2\right)\omega_{shell}^2 = \frac{1}{3}MR^2\left(\frac{v}{R}\right)^2.\]
06

Calculate Energy Expressions

Substitute \(I_{shell}\) and the expressions for \(\omega\) into the final kinetic energy:\[KE_f = \frac{1}{2}mv^2 + \frac{1}{3}Mv^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2.\]
07

Set up Conservation Equation

With conservation of energy, \(PE_i = KE_f\):\[0.60 \times 9.8 \times 0.82 = \frac{1}{2} \times 0.60 \times v^2 + \frac{1}{3}\times 4.5 \times v^2 + \frac{1}{2}\times 3.0 \times 10^{-3}\times \left(\frac{v}{0.05}\right)^2.\]
08

Solve for v

Calculate \(v\) by solving the equation:Simplifying, \[4.788 = 0.30v^2 + 0.833v^2 + 1.2v^2,\]which leads to \[v^2 \approx \frac{4.788}{2.333}.\]Finally, solve for \(v\): \[v \approx \sqrt{2.05} \approx 1.43 \mathrm{\,m/s}.\]

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Key Concepts

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

Kinetic Energy
Kinetic energy is a crucial concept when studying motion and energy conservation. When an object is in motion, it carries energy due to its velocity, known as kinetic energy. The kinetic energy of a moving object depends on both its mass and its speed. The formula is expressed as:
  • \( KE = \frac{1}{2}mv^2 \)
where \(m\) is the mass of the object, and \(v\) is its velocity.
This concept is essential because it helps us understand how energy is conserved in systems involving movement. For the object falling in the given exercise, its potential energy as it starts from rest is converted into kinetic energy as it gains speed while falling.
This transformation follows the law of conservation of energy, showing how energy can transfer from one form to another without being lost.
In the context of the given problem, not only does the small object gain translational kinetic energy, but the sphere and the pulley also gain rotational kinetic energy. Each component of the system contributes to the total energy, helping us calculate the speed.
Rotational Motion
Rotational motion occurs when an object rotates around an axis. It's a key concept in understanding how different parts of a system move and interact, especially when dealing with complex systems like the spherical shell and pulley in the exercise.
When comparing rotational motion to linear motion, rotational kinematics deal with angles and angular velocity (\( \omega \)). For example, in this exercise, the pulley and the shell both rotate, and this affects the translational motion of the object falling.
  • The connection between linear and angular motion through the equations: \(v = r\omega\), where \(v\) is linear velocity and \( r \) is the radius of rotation.
  • For the shell, the relationship \(v = R\omega_{shell}\) shows how the speed of the object affects the spherical shell's rotation.
Since \( \omega_{shell} = \frac{v}{R} \), as the small object moves, it causes corresponding angular motion in the shell and pulley. Understanding these connections helps solve problems involving rotational and translational velocity, which is essential for finding the system's final speed (\(v\)).
Spherical Shell
A spherical shell is an important concept in physics, particularly when dealing with rotational dynamics. It is essentially a hollow, spherical object where mass is distributed uniformly across its surface. In this exercise, the spherical shell is part of a rotational system that includes a pulley and a falling mass.
The spherical shell influences the dynamics due to its unique geometry and distribution of mass.
  • When it rotates, each part of the shell moves in a circular path, contributing to rotational inertia.
  • Moment of inertia for a spherical shell is calculated as: \( I_{shell} = \frac{2}{3}MR^2 \), where \(M\) is the mass of the shell and \(R\) is its radius.
The spherical shell's role in the problem is to convert part of the falling object's potential energy into rotational energy. The energy dynamics are divided among the linear kinetic energy of the object, the rotational kinetic energy of the shell, and the rotational energy of the pulley.
Understanding the spherical shell's role helps us appreciate how energy gets distributed in systems involving rotation. By knowing how its inertia affects the system, we can accurately predict the speed of the object in motion.
Moment of Inertia
Moment of inertia is a fundamental principle in rotational dynamics. It measures an object's resistance to changes in its rotational motion. It depends on how an object's mass is distributed relative to the axis of rotation. In this problem, both the spherical shell and the pulley have a moment of inertia, influencing the system's overall motion.
  • For a point mass, moment of inertia is calculated as: \( I = mr^2 \), where \(m\) is the mass and \(r\) is the distance from the rotation axis.
  • For distributed masses, like the spherical shell, specific formulas account for the entire shape's mass distribution.
In this system, the moment of inertia is critical in the energy calculations. Inertia determines how the rotational energy is shared between the pulley and the spherical shell.
The larger the moment of inertia, the more resistant an object is to changes in its rotational speed. For the spherical shell and pulley:
  • Pulley inertia: \(I_{pulley} = 3.0 \times 10^{-3} \text{ kg} \cdot \text{m}^2\).
  • Shell inertia: \(I_{shell} = \frac{2}{3} \times 4.5 \times (0.085)^2 \).
These inertial values help us understand how much rotational motion is involved, completing the picture of how energy conservation leads to the object's final velocity.

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Most popular questions from this chapter

A horizontal pottery wheel (a horizontal disk) with a radius of \(30.0 \mathrm{~cm}\) can rotate about a vertical axis with negligible friction but is initially stationary. A horizontal rubber wheel of radius \(2.00 \mathrm{~cm}\) is placed against its rim. That wheel is mounted on a motor. When the motor is switched on at time \(t=0\), the rubber wheel undergoes a constant angular acceleration of \(5.00 \mathrm{rad} / \mathrm{s}^{2}\). Its contact with the pottery wheel causes the pottery wheel to undergo an angular acceleration. When the pottery wheel reaches an angular speed of \(5.00 \mathrm{rev} / \mathrm{s}\), the rubber wheel is pulled away from contact and thereafter the pottery wheel rotates at \(5.00 \mathrm{rev} / \mathrm{s}\). From \(t=0\) to \(t=2.00 \mathrm{~min}\), how many full rotations does the pottery wheel make?

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