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Chapter 9: Problem 43

Wheel \(A\) has three times the moment of inertia about its axis of rotation as wheel \(B .\) Wheel \(B\) 's angular speed is four times that of wheel \(A\). (a) Which wheel has the greater rotational kinetic energy? (b) If \(K_{A}\) and \(K_{B}\) are the rotational kinetic energies of the wheels, what is \(K_{A} / K_{B} ?\)

Short Answer

Expert verified
Wheel \(B\) has greater rotational kinetic energy. The ratio \(K_{A} / K_{B} = \frac{3}{4}\).

Step by step solution

01

Read and Understand the Problem

In the problem, two wheels \(A\) and \(B\) are given. The moment of inertia of wheel \(A\) is three times that of wheel \(B\) denoted as \(I_{A} = 3I_{B}\). Wheel \(B\) 's angular speed is four times that of wheel \(A\) denoted as \(\omega_{B} = 4\omega_{A}\). We are asked to find which wheel has the greater rotational kinetic energy and the value of \(K_{A} / K_{B}\).
02

Formulate Equations for Kinetic Energy

Begin by using the equation of rotational kinetic energy \(K = \frac{1}{2} I \omega^2\). Substituting in the relations between moments of inertia and angular speeds between the two wheels, we get \(K_{A} = \frac{1}{2} 3I_{B} \omega_A^2\) and \(K_{B} = \frac{1}{2} I_{B} (4 \omega_{A})^2\). Simplify these to get \(K_{A} = \frac{3}{2} I_{B} \omega_{A}^2\) and \(K_{B} = 2 I_{B} \omega_{A}^2\).
03

Determine Which Wheel Has Greater Kinetic Energy

Just by comparing \(K_{A}\) and \(K_{B}\), it can be seen that \(K_{B}\) is larger than \(K_{A}\) because \(2 I_{B} \omega_{A}^2 > \frac{3}{2} I_{B} \omega_{A}^2\). Thus, wheel \(B\) has greater kinetic energy.
04

Compute the Ratio \(K_{A} / K_{B}\)

Calculate the ratio \(K_{A} / K_{B} = (\frac{3}{2} I_{B} \omega_{A}^2) / (2 I_{B} \omega_{A}^2) = \frac{3}{4}\).

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Key Concepts

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

Moment of Inertia
Moment of inertia, often represented by the symbol 'I', is a measure of an object's resistance to changes in its rotation rate. It's an inertial property of a rigid body that quantifies how the distribution of mass is spaced relative to an axis of rotation. Just like mass is a measure of how much an object resists linear acceleration, moment of inertia is that measure for rotational acceleration.

In the given exercise, wheel A has a moment of inertia which is three times that of wheel B, mathematically represented as \(I_{A} = 3I_{B}\). This fundamental concept dictates how much torque is required for a wheel to reach a certain angular speed, or simply put, how 'hard' it is to spin the wheel. The greater the moment of inertia, the more energy it will take to change the wheel's rotational speed.
Angular Speed
Angular speed, denoted by the symbol '\(\omega\)', refers to how fast an object rotates or revolves relative to another point, which is typically the object's center of rotation. It's analogous to linear speed but instead applies to rotational movement. Angular speed is measured in radians per second (rad/s).

In our exercise, wheel B's angular speed is four times that of wheel A (\(\omega_{B} = 4\omega_{A}\)). The equation \(K = \frac{1}{2} I \omega^2\) relates angular speed to rotational kinetic energy, demonstrating that the energy is not only dependent on the speed itself but also the square of it. This means that even a slight increase in angular speed can greatly increase the rotational kinetic energy.
Physics Problem Solving
Effective physics problem solving typically involves the clear understanding of concepts and formulas, identifying relevant information, and methodically applying logical steps to find the solution. In the context of this exercise, it primarily involved identifying the relationship between moment of inertia and angular speed as they contribute to rotational kinetic energy.

By breaking down the problem into smaller steps, one can systematically determine which wheel has greater kinetic energy and the exact ratio of their kinetic energies. Knowing the right equations to use and how to manipulate them plays a crucial role. The step-by-step solution provided demonstrates a logical progression from reading and understanding the problem, through to computing the final answer, which is a quintessential skill set in physics problem solving.

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Most popular questions from this chapter

Two metal disks, one with radius \(R_{1}=2.50 \mathrm{~cm}\) and mass \(M_{1}=0.80 \mathrm{~kg}\) and the other with radius \(R_{2}=5.00 \mathrm{~cm}\) and mass \(M_{2}=1.60 \mathrm{~kg},\) are welded together and mounted on a frictionless axis through their common center (Fig. \(\mathbf{P 9 . 7 9}\) ). (a) What is the total moment of inertia of the two disks? (b) A light string is wrapped around the edge of the smaller disk, and a \(1.50 \mathrm{~kg}\) block is suspended from the free end of the string. If the block is released from rest at a distance of \(2.00 \mathrm{~m}\) above the floor, what is its speed just before it strikes the floor? (c) Repeat part (b), this time with the string wrapped around the edge of the larger disk. In which case is the final speed of the block greater? Explain.

At \(t=3.00 \mathrm{~s}\) a point on the rim of a \(0.200-\mathrm{m}\) -radius wheel has a tangential speed of \(50.0 \mathrm{~m} / \mathrm{s}\) as the wheel slows down with a tangential acceleration of constant magnitude \(10.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) Calculate the wheel's constant angular acceleration. (b) Calculate the angular velocities at \(t=3.00 \mathrm{~s}\) and \(t=0 .\) (c) Through what angle did the wheel turn between \(t=0\) and \(t=3.00 \mathrm{~s} ?\) (d) At what time will the radial acceleration equal \(g ?\)

A circular saw blade with radius \(0.120 \mathrm{~m}\) starts from rest and turns in a vertical plane with a constant angular acceleration of \(2.00 \mathrm{rev} / \mathrm{s}^{2} .\) After the blade has turned through \(155 \mathrm{rev},\) a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of \(0.820 \mathrm{~m}\) to the floor. How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?

A bicycle wheel has an initial angular velocity of \(1.50 \mathrm{rad} / \mathrm{s}\) (a) If its angular acceleration is constant and equal to \(0.200 \mathrm{rad} / \mathrm{s}^{2},\) what is its angular velocity at \(t=2.50 \mathrm{~s} ?\) (b) Through what angle has the wheel turned between \(t=0\) and \(t=2.50 \mathrm{~s} ?\)

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant linear speed of \(v=1.25 \mathrm{~m} / \mathrm{s} .\) Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the \(\mathrm{CD}\) is played. (See Exercise \(9.20 .\) ) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta)=r_{0}+\beta \theta,\) where \(r_{0}\) is the radius of the spiral at \(\theta=0\) and \(\beta\) is a constant. On a \(\mathrm{CD}, r_{0}\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d \theta,\) the distance scanned along the track is \(d s=r d \theta .\) Using the above expression for \(r(\theta),\) integrate \(d s\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) since the track is scanned at a constant linear speed \(v,\) the distance \(s\) found in part (a) is equal to vi. Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta ;\) choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_{z}\) and the angular acceleration \(\alpha_{z}\) as functions of time. Is \(\alpha_{z}\) constant? (d) On a CD, the inner radius of the track is \(25.0 \mathrm{~mm}\), the track radius increases by \(1.55 \mu \mathrm{m}\) per revolution, and the playing time is \(74.0 \mathrm{~min} .\) Find \(r_{0}, \beta,\) and the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_{z}\) (in rad/s) versus \(t\) and \(\alpha_{z}\) (in rad/s \(^{2}\) ) versus \(t\) between \(t=0\) and \(t=74.0 \mathrm{~min}\)
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