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Chapter 8: Problem 60

A hollow cylinder, a uniform solid sphere, and a uniform solid cylinder all have the same mass \(m .\) The three objects are rolling on a horizontal surface with identical transnational speeds \(v .\) Find their total kinetic energies in terms of \(m\) and \(v\) and order them from smallest to largest.

Short Answer

Expert verified
Question: Arrange the following objects in order of increasing total kinetic energy: a hollow cylinder, a uniform solid sphere, and a uniform solid cylinder, all having the same mass and rolling on a horizontal plane without slipping at the same constant translational speed. Answer: Uniform solid sphere < Uniform solid cylinder < Hollow cylinder

Step by step solution

01

Calculate the translational kinetic energy

As all objects have the same mass and speed, we can calculate their translational kinetic energy as: \(K_t = \frac{1}{2}mv^2\)
02

Calculate the rotational kinetic energy for each object

Using each object's moment of inertia and the relationship between linear and angular velocity, we can calculate their rotational kinetic energy as follows: 1. Hollow cylinder: \(K_{r1} =\frac{1}{2}I_1\omega^2= \frac{1}{2}(mr^2)\left(\frac{v^2}{r^2}\right) = \frac{1}{2}mv^2\) 2. Uniform solid sphere: \(K_{r2} = \frac{1}{2}I_2\omega^2= \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v^2}{r^2}\right) = \frac{1}{5}mv^2\) 3. Uniform solid cylinder: \(K_{r3} = \frac{1}{2}I_3\omega^2= \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v^2}{r^2}\right) = \frac{1}{4}mv^2\)
03

Calculate the total kinetic energy for each object

Add the translational and rotational kinetic energy for each object: 1. Hollow cylinder: \(K_1 = K_t + K_{r1} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2\) 2. Uniform solid sphere: \(K_2 = K_t + K_{r2} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2\) 3. Uniform solid cylinder: \(K_3 = K_t + K_{r3} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2\)
04

Order the objects from smallest to largest total kinetic energy

Comparing the total kinetic energies calculated in step 3, we find the following order: 1. Uniform solid sphere: \(K_2 = \frac{7}{10}mv^2\) 2. Uniform solid cylinder: \(K_3 = \frac{3}{4}mv^2\) 3. Hollow cylinder: \(K_1 = mv^2\) So, the total kinetic energies are ordered as: Uniform solid sphere < Uniform solid cylinder < Hollow cylinder.

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Most popular questions from this chapter

A hoop of \(2.00-\mathrm{m}\) circumference is rolling down an inclined plane of length \(10.0 \mathrm{m}\) in a time of \(10.0 \mathrm{s} .\) It started out from rest. (a) What is its angular velocity when it arrives at the bottom? (b) If the mass of the hoop, concentrated at the rim, is \(1.50 \mathrm{kg},\) what is the angular momentum of the hoop when it reaches the bottom of the incline? (c) What force(s) supplied the net torque to change the hoop's angular momentum? Explain. [Hint: Use a rotation axis through the hoop's center. \(]\) (d) What is the magnitude of this force?
Verify that the units of the rotational form of Newton's second law [Eq. (8-9)] are consistent. In other words, show that the product of a rotational inertia expressed in \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and an angular acceleration expressed in \(\mathrm{rad} / \mathrm{s}^{2}\) is a torque expressed in \(\mathrm{N} \cdot \mathrm{m}\).
Find the force exerted by the biceps muscle in holding a 1-L milk carton (weight \(9.9 \mathrm{N}\) ) with the forearm parallel to the floor. Assume that the hand is \(35.0 \mathrm{cm}\) from the elbow and that the upper arm is $30.0 \mathrm{cm}$ long. The elbow is bent at a right angle and one tendon of the biceps is attached to the forearm at a position \(5.00 \mathrm{cm}\) from the elbow, while the other tendon is attached at \(30.0 \mathrm{cm}\) from the elbow. The weight of the forearm and empty hand is \(18.0 \mathrm{N}\) and the center of gravity of the forearm is at a distance of \(16.5 \mathrm{cm}\) from the elbow.
The angular momentum of a spinning wheel is $240 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} .$ After application of a constant braking torque for \(2.5 \mathrm{s}\), it slows and has a new angular momentum of $115 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} .$ What is the torque applied?
A bicycle wheel, of radius \(0.30 \mathrm{m}\) and mass \(2 \mathrm{kg}\) (concentrated on the rim), is rotating at 4.00 rev/s. After 50 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces?
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