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A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of 0.890 rad/s\(^2\). What is the kinetic energy of the shell after it has turned through 6.00 rev?

Short Answer

Expert verified
The kinetic energy of the shell is approximately 8.88 J after 6 revolutions.

Step by step solution

01

Calculate Angular Displacement

Convert the number of revolutions to radians. Since 1 revolution is \(2\pi\) radians, \\[ 6.00 \text{ rev} \times 2\pi \text{ radians/rev} = 12\pi \text{ radians}. \]
02

Use Equation for Final Angular Velocity

Use the equation for angular motion given constant angular acceleration: \\[ \omega^2 = \omega_0^2 + 2\alpha\theta, \] \where \(\omega_0 = 0\) (initial angular velocity), \(\alpha = 0.890 \text{ rad/s}^2\), and \(\theta = 12\pi \text{ radians}\). \Thus, \\[ \omega = \sqrt{0 + 2(0.890)(12\pi)}. \]
03

Solve for Final Angular Velocity

Calculate the final angular velocity, \(\omega\): \\[ \omega = \sqrt{2(0.890)(12\pi)} \approx \sqrt{67.092} \approx 8.19 \text{ rad/s}. \]
04

Use Moment of Inertia for Hollow Spherical Shell

The moment of inertia \(I\) for a hollow spherical shell is given by \\[ I = \frac{2}{3}mr^2, \] \where \(m = 8.20 \text{ kg}\) and \(r = 0.220 \text{ m}\). \Thus, \\[ I = \frac{2}{3}(8.20)(0.220)^2. \]
05

Calculate Moment of Inertia

Substitute the values into the moment of inertia formula: \\[ I = \frac{2}{3}(8.20)(0.220)^2 = \frac{2}{3}(8.20)(0.0484) = 0.2647 \text{ kg}\cdot\text{m}^2. \]
06

Calculate Kinetic Energy

The rotational kinetic energy \(KE\) is given by \\[ KE = \frac{1}{2}I\omega^2. \] \Substitute \(I = 0.2647 \text{ kg}\cdot\text{m}^2\) and \(\omega = 8.19 \text{ rad/s}\): \\[ KE = \frac{1}{2}(0.2647)(8.19)^2 \approx \frac{1}{2}(0.2647)(67.0921) \approx 8.88 \text{ J}. \]

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Key Concepts

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

Angular Displacement
Angular displacement refers to how much an object has rotated or turned over a certain period. It's measured in radians, a standard unit in rotational motion. To understand it deeply, picture a circle. One complete trip around the circle, or one revolution, equals 2\(\pi\) radians. When you know how many revolutions an object has gone through, you can find its angular displacement by simply multiplying the number of revolutions by 2\(\pi\). In the original exercise, a hollow spherical shell is initially at rest and later rotates by six full revolutions. To find the angular displacement, multiply 6 by 2\(\pi\), resulting in 12\(\pi\) radians. This value tells you the total angle through which the shell has turned. This understanding is crucial when calculating other properties of rotational motion, such as angular velocity and final kinetic energy.
Angular Velocity
Angular velocity defines how fast an object rotates or spins around an axis. It's akin to linear velocity but specifically applied to circular paths or motions. Angular velocity is expressed in radians per second (rad/s). When an object is under constant angular acceleration, like the hollow spherical shell in our exercise, you can find its final angular velocity using the formula: \[ \omega^2 = \omega_0^2 + 2\alpha\theta, \]where \(\omega_0\) is the initial angular velocity (zero if the object starts from rest), \(\alpha\) is the angular acceleration, and \(\theta\) is the angular displacement.For our exercise, start from rest (\(\omega_0 = 0\)), with \(\alpha = 0.890 \ \text{rad/s}^2\) and \(\theta = 12\pi \ \text{radians}\). Substituting these into the formula, you find the final angular velocity \(\omega\) to be approximately 8.19 rad/s. Knowing angular velocity helps you understand how swiftly or gently the object is spinning after a specific displacement or time.
Moment of Inertia
Moment of inertia is a property that measures an object's resistance to changes in its rotational motion. It's somewhat similar to mass in linear motion but specifically for rotation. The bigger the moment of inertia, the harder it is to change the object's state of rotation. For different shapes, you compute the moment of inertia differently. For a hollow spherical shell, the formula is: \[ I = \frac{2}{3}mr^2, \]where \(m\) is the mass and \(r\) is the radius.In our problem, the hollow spherical shell has a mass of 8.20 kg and a radius of 0.220 m. By substituting these values into the formula, we get a moment of inertia \(I\) of 0.2647 kg\(\cdot\)m\(^2\). This calculation is pivotal because it directly influences how much rotational kinetic energy your object will possess once it's in motion.

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Most popular questions from this chapter

A computer disk drive is turned on starting from rest and has constant angular acceleration. If it took 0.0865 s for the drive to make its \(second\) complete revolution, (a) how long did it take to make the first complete revolution, and (b) what is its angular acceleration, in rad/s\(^2\)?

About what axis will a uniform, balsa-wood sphere have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius, with the axis along a diameter?

The Crab Nebula is a cloud of glowing gas about 10 lightyears across, located about 6500 light-years from the earth (\(\textbf{Fig. P9.86}\)). It is the remnant of a star that underwent a \(supernova\) \(explosion\), seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about 5 \(\times\) 10\(^3$$^1\) W, about 10\(^5\) times the rate at which the sun radiates energy. The Crab Nebula obtains its energy from the rotational kinetic energy of a rapidly spinning \(neutron\) \(star\) at its center. This object rotates once every 0.0331 s, and this period is increasing by 4.22 \(\times 10^{-13}\) s for each second of time that elapses. (a) If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star. (b) Theories of supernovae predict that the neutron star in the Crab Nebula has a mass about 1.4 times that of the sun. Modeling the neutron star as a solid uniform sphere, calculate its radius in kilometers. (c) What is the linear speed of a point on the equator of the neutron star? Compare to the speed of light. (d) Assume that the neutron star is uniform and calculate its density. Compare to the density of ordinary rock (3000 kg/m\(^3\)) and to the density of an atomic nucleus (about 10\(^{17}\) kg/m\(^3\)). Justify the statement that a neutron star is essentially a large atomic nucleus.

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

Small blocks, each with mass \(m\), are clamped at the ends and at the center of a rod of length \(L\) and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point one-fourth of the length from one end.

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