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

Short Answer

Expert verified
The kinetic energy of the spherical shell is approximately 57.15 J.

Step by step solution

01

Determine Angular Displacement in Radians

First, convert the number of revolutions to radians, since 1 revolution is equal to \(2\pi\) radians. Thus, 6.00 revolutions is equivalent to \(6.00 \times 2\pi = 12\pi\) radians.
02

Calculate Final Angular Velocity

Use the kinematic equation for angular motion, \(\omega^2 = \omega_0^2 + 2\alpha\theta\), where \(\omega_0 = 0\), \(\alpha = 0.890\, \mathrm{rad/s^2}\), and \(\theta = 12\pi\, \mathrm{rad}\). Substitute these values into the equation: \[\omega^2 = 0 + 2 \times 0.890 \times 12\pi = 21.36\pi\] Solve for \(\omega\):\[\omega = \sqrt{21.36\pi}\]
03

Use Moment of Inertia for Kinetic Energy

The moment of inertia \(I\) for a hollow sphere is \(I = \frac{2}{3}MR^2\). With \(M = 8.20\, \mathrm{kg}\) and \(R = 0.220\, \mathrm{m}\), calculate \(I\): \[I = \frac{2}{3} \times 8.20 \times 0.220^2 = 0.2651\, \mathrm{kg\,m^2}\]
04

Calculate Kinetic Energy

The rotational kinetic energy \(KE\) is given by \(\frac{1}{2}I\omega^2\). Plug in the values for \(I\) and \(\omega\) to compute \(KE\): \[KE = \frac{1}{2} \times 0.2651 \times (\sqrt{21.36\pi})^2\] Simplify and calculate the value of \(KE\).

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Key Concepts

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

Angular Displacement
In rotational kinematics, **angular displacement** refers to the angle through which an object rotates around a particular axis. It is the rotational equivalent of linear displacement and is typically measured in radians. Think of it as the angular "distance" the object covers during its rotation.

Angular displacement is fundamental in calculations involving rotational motion. To better understand, recall that one complete revolution around a circle corresponds to an angular displacement of \(2\pi\) radians. In our exercise, we deal with 6 revolutions, which equate to an angular displacement of \(6 \times 2\pi = 12\pi\) radians.

This transformation from revolutions to radians is common practice in physics, as radians provide a convenient way to relate angular quantities to linear ones. By knowing the angular displacement, we can proceed to use other rotational motion formulas to find more information about the system's state, such as its angular velocity.
Moment of Inertia
The **moment of inertia** is a measure of an object's resistance to change in its rotational motion. It's somewhat analogous to mass in linear motion, but instead of measuring 'how much' matter an object has, it measures 'how far' the mass is distributed from the axis of rotation.

For an object like a hollow sphere, the calculation of the moment of inertia is governed by the formula:
  • \(I = \frac{2}{3}MR^2\)
Where \(M\) is the mass, and \(R\) is the radius of the sphere. By substituting the mass (8.20 kg) and radius (0.220 m) into this formula, we determine that the moment of inertia \(I\) is 0.2651 kg m².

Understanding moment of inertia is crucial when dealing with rotational dynamics, as it directly affects the rotational kinetic energy and angular acceleration and plays a critical role when applying Newton's second law in rotational form. It tells us how much torque is required to achieve a desired angular acceleration.
Rotational Kinetic Energy
**Rotational kinetic energy** is the energy possessed by a rotating object due to its motion. It's important in understanding the dynamics of systems involving rotation. The formula for rotational kinetic energy \(KE\) shares similarities with linear kinetic energy:
  • \(KE = \frac{1}{2}I\omega^2\)
Where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

In our case, after calculating \(I = 0.2651 \text{ kg m}^2\) and the angular velocity \(\omega\), we insert these values into the formula to calculate the rotational kinetic energy. This energy quantifies how effectively the sphere converts its rotational motion into energy.

Rotational kinetic energy is an essential concept, particularly when understanding the conversion of energy forms, such as potential energy into kinetic energy during a rolling motion. This also has practical applications in the analysis of mechanical systems where rotational motion is a factor.

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Most popular questions from this chapter

A disk of radius 25.0 \(\mathrm{cm}\) cm is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. P9.65). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is 1.80 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) Find \(A .\) (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .\) .

A sphere consists of a solid wooden ball of uniform density 800 \(\mathrm{kg} / \mathrm{m}^{3}\) and radius 0.30 \(\mathrm{m}\) and is covered with a thin coating of lead foil with area density 20 \(\mathrm{kg} / \mathrm{m}^{2}\) . Calculate the moment of inertia of this sphere about an axis passing through its center.

It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball bearings. Consider a flywheel made of iron (density 7800 \(\mathrm{kg} / \mathrm{m}^{3}\) ) in the shape of a 10.0 -cm-thick uniform disk. (a) What would the diameter of such a disk need to be if it is to store 10.0 megajoules of kinetic energy when spinning at 90.0 rpm about an axis perpendicular to the disk at its center? (b) What would be the centripetal acceleration of a point on its rim when spinning at this rate?

A thin, light wire is wrapped around the rim of a wheel, as shown in Fig. E9.49. The wheel rotates without friction about a stationary horizontal axis that passes through the center of the wheel. The wheel is a uniform disk with radius \(R=0.280 \mathrm{m}\) . An object of mass \(m=4.20 \mathrm{kg}\) is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration. If the suspended object moves downward a distance of 3.00 \(\mathrm{m}\) in \(2.00 \mathrm{s},\) what is the mass of the wheel?

An airplane propeller is rotating at 1900 rpm (rev/min). (a) Compute the propeller's angular velocity in rad/s. (b) How many seconds does it take for the propeller to turn through \(35^{\circ} ?\)

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