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Chapter 10: Problem 17

A uniform solid sphere of radius \(R, \operatorname{mass} M,\) and moment of inertia \(I=\frac{2}{5} M R^{2}\) is rolling without slipping along a horizontal surface. Its total kinetic energy is the sum of the energies associated with translation of the center of mass and rotation about the center of mass. Find the fraction of the sphere's total kinetic energy that is attributable to rotation.

Short Answer

Expert verified
The fraction of the sphere's total kinetic energy that is due to rotation is 2/7.

Step by step solution

01

Find the translational kinetic energy

To find the translational kinetic energy, we use the formula: \(T = \frac{1}{2}mv^2\) where \(m\) is the mass of the sphere, and \(v\) is its linear velocity.
02

Find the rotational kinetic energy

To find the rotational kinetic energy, we use the formula: \(R = \frac{1}{2}I\omega^2\) where \(I\) is the moment of inertia of the sphere, and \(\omega\) is its angular velocity.
03

Relate the linear and angular velocities

Since the sphere is rolling without slipping, we can relate the linear velocity \(v\) to its angular velocity \(\omega\): \(v = R\omega\)
04

Express the rotational kinetic energy in terms of linear velocity

Substitute the relation between linear and angular velocities into the expression for rotational kinetic energy: \(R = \frac{1}{2}I(\frac{v}{R})^2 = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2}) = \frac{1}{5}mv^2\)
05

Find the total kinetic energy

The total kinetic energy is the sum of translational and rotational kinetic energies: \(K = T + R = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2\)
06

Calculate the fraction of rotational kinetic energy in the total kinetic energy

Finally, we find the fraction of the rotational kinetic energy in the total kinetic energy: \(\frac{R}{K} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{1}{5}\cdot\frac{10}{7} = \boxed{\frac{2}{7}}\) So, the fraction of the sphere's total kinetic energy that is attributable to rotation is \(\frac{2}{7}\).

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Most popular questions from this chapter

A uniform solid cylinder of mass \(M=5.00 \mathrm{~kg}\) is rolling without slipping along a horizontal surface. The velocity of its center of mass is \(30.0 \mathrm{~m} / \mathrm{s}\). Calculate its energy.

A uniform solid sphere of mass \(m\) and radius \(r\) is placed on a ramp inclined at an angle \(\theta\) to the horizontal. The coefficient of static friction between sphere and ramp is \(\mu_{s} .\) Find the maximum value of \(\theta\) for which the sphere will roll without slipping, starting from rest, in terms of the other quantities.

A solid sphere of radius \(R\) and mass \(M\) is placed at a height \(h_{0}\) on an inclined plane of slope \(\theta\). When released, it rolls without slipping to the bottom of the incline. Next, a cylinder of same mass and radius is released on the same incline. From what height \(h\) should it be released in order to have the same speed as the sphere at the bottom?

A sheet of plywood \(1.3 \mathrm{~cm}\) thick is used to make a cabinet door \(55 \mathrm{~cm}\) wide by \(79 \mathrm{~cm}\) tall, with hinges mounted on the vertical edge. A small 150 - \(\mathrm{g}\) handle is mounted \(45 \mathrm{~cm}\) from the lower hinge at the same height as that hinge. If the density of the plywood is \(550 \mathrm{~kg} / \mathrm{m}^{3},\) what is the moment of inertia of the door about the hinges? Neglect the contribution of hinge components to the moment of inertia.

The flywheel of an old steam engine is a solid homogeneous metal disk of mass \(M=120 . \mathrm{kg}\) and radius \(R=80.0 \mathrm{~cm} .\) The engine rotates the wheel at \(500 .\) rpm. In an emergency, to bring the engine to a stop, the flywheel is disengaged from the engine and a brake pad is applied at the edge to provide a radially inward force \(F=100 .\) N. If the coefficient of kinetic friction between the pad and the flywheel is \(\mu_{\mathrm{k}}=0.200,\) how many revolutions does the flywheel make before coming to rest? How long does it take for the flywheel to come to rest? Calculate the work done by the torque during this time.
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