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Chapter 9: Problem 50

A bucket of mass \(m\) is tied to a massless cable that is wrapped around the outer rim of a frictionless uniform pulley of radius \(R\) , similar to the system shown in Fig. \(9.32 .\) In terms of the stated variables, what must be the moment of inertia of the pulley so that it always has half as much kinetic energy as the bucket?

Short Answer

Expert verified
The moment of inertia of the pulley is \( \frac{1}{2} m R^2 \).

Step by step solution

01

Understand the Problem

We are given a bucket with mass \(m\) tied to a massless cable around a frictionless pulley with radius \(R\). The problem asks us to find the moment of inertia \(I\) of the pulley so that it has half the kinetic energy of the bucket in terms of the given variables.
02

Set the Known Relationships

The kinetic energy of the bucket, which is moving linearly, is given by \( K_{b} = \frac{1}{2} m v^2 \), where \( v \) is the velocity. The kinetic energy of the pulley, which is rotating, is \( K_{p} = \frac{1}{2} I \omega^2 \). Given, \( K_{p} = \frac{1}{2} K_{b} \).
03

Express Angular Velocity

Since linear velocity \( v \) and angular velocity \( \omega \) are related by the radius \( R \) of the pulley, we use \( v = R \omega \). Hence, \( \omega = \frac{v}{R} \).
04

Substitute Relationships in Kinetic Energy

Substitute the expression for \( \omega \) into the kinetic energy of the pulley to make it in terms of \( v \): \( K_{p} = \frac{1}{2} I \left(\frac{v}{R}\right)^2 = \frac{1}{2} I \frac{v^2}{R^2} \).
05

Set up the Kinetic Energy Equation

Setting \( K_{p} = \frac{1}{2} K_{b} \), we get \( \frac{1}{2} I \frac{v^2}{R^2} = \frac{1}{2} \left(\frac{1}{2} m v^2\right) \). Simplifying gives \( \frac{1}{2} I \frac{v^2}{R^2} = \frac{1}{4} m v^2 \).
06

Solve for Moment of Inertia \(I\)

From the equation \( \frac{1}{2} I \frac{v^2}{R^2} = \frac{1}{4} m v^2 \), cancel out \( v^2 \) on both sides and solve for \( I \): \( \frac{1}{2} \frac{I}{R^2} = \frac{1}{4} m \), which simplifies to \( I = \frac{1}{2} m R^2 \).

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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 fundamental concept in physics that describes the energy of an object in motion. For any object with mass, if it's moving, it possesses kinetic energy. This energy is dependent both on the mass of the object and its velocity. The formula for kinetic energy is given by \[ K = \frac{1}{2} m v^2 \]where:
  • \( K \) is the kinetic energy,
  • \( m \) is the mass, and
  • \( v \) is the velocity of the object.
In this exercise, we have two sources of kinetic energy: the linearly moving bucket and the rotating pulley. The bucket鈥檚 kinetic energy differs from the pulley鈥檚 due to its linear motion, calculated directly using its mass and linear velocity. Understanding this helps us connect how energy impacts different forms of motion based on velocity.
Linear Velocity
Linear velocity represents the rate of change of an object's position over time, in a straight line. It's a vector quantity, meaning it has both magnitude and direction. For the bucket in the exercise, its linear velocity is crucial as its motion dictates many parts of the system's behavior. The speed of the bucket can be denoted as:\[ v = R \omega \]where \( \omega \) is the angular velocity of the pulley. Here, linear velocity and angular velocity are interdependent, affecting each other proportionally depending on the radius \( R \). When considering systems involving pulleys, ropes, or other mechanics, understanding linear velocity helps clarify how movement translates throughout a system, particularly in energy calculations.
Angular Velocity
Angular velocity describes how fast an object is rotating, denoted by \( \omega \), and is crucial in understanding rotary motion. It connects directly to linear velocity. The formula:\[ \omega = \frac{v}{R} \]links it to the linear velocity \( v \) of the bucket and the radius \( R \) of the pulley. Angular velocity is essential when assessing the kinetic energy of rotating systems, like the pulley in this exercise. This relationship underscores how rotational dynamics can be understood in terms of more familiar linear dynamics. To calculate the pulley鈥檚 kinetic energy, you use the angular velocity and moment of inertia, emphasizing the interconnectedness of these physical properties.

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Most popular questions from this chapter

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

A thin, rectangular sheet of metal has mass \(M\) and sides of length \(a\) and \(b\) . Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one comer of the sheet.

An electric turntable 0.750 \(\mathrm{m}\) in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 rev/s and a constant angular acceleration of 0.900 rev \(/ \mathrm{s}^{2} .\) (a) Compute the angular velocity of the turntable after 0.200 \(\mathrm{s}\) . (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200 \mathrm{s} ?(\mathrm{d})\) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{s} ?\)

A wheel is rotating about an axis that is in the \(z\) -direction. The angular velocity \(\omega_{z}\) is \(-6.00 \mathrm{rad} / \mathrm{s}\) at \(t=0,\) increases linearly with time, and is \(+8.00 \mathrm{m} / \mathrm{s}\) at \(t=7.00 \mathrm{s}\) . We have taken counterclockwise rotation to be positive. (a) Is the angular acceleration during this time interval positive or negative? (b) During what time interval is the speed of the wheel increasing? Decreasing? (c) What is the angular displacement of the wheel at \(t=7.00 \mathrm{s} ?\)

Calculate the moment of inertia of a uniform solid cone about an axis through its center (Fig. 9.40). The cone has mass \(M\) and altitude \(h\) . The radius of its circular base is \(R\) . figure can't copy
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