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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 rev?

Short Answer

Expert verified
The kinetic energy of the shell after it has turned through 6.00 revolution is calculated by substituting the appropriate values into the kinetic energy formula.

Step by step solution

01

Find the Total Rotation Angle

To solve this exercise, first calculate the total angle turned through by the shell. Since the shell has turned through 6 revolutions, and each revolution is equal to \(2\pi\) radians, therefore the total angle \(\alpha = 6.00\cdot 2\pi\).
02

Calculate the Final Angular Speed

Having identified the total angle of rotation, calculate the final angular speed using the equation of rotational motion \( \omega^2 = \omega_0^2 + 2\cdot \alpha\cdot a \), where \(\omega_0\) is initial angular velocity, \(\alpha\) is angular displacement and \(a\) is angular acceleration. Since the shell initially was at rest, \( \omega_0 = 0 \). Substituting the values we have, we can solve for the final angular speed \( \omega \).
03

Compute the Rotational Kinetic Energy

Finally, given the mass and the radius of the shell, we can calculate the moment of inertia, \( I \), of the hollow spherical shell. The formula for the moment of inertia for a hollow sphere rotating about a diameter is \( I =\frac{2}{3}mr^2 \). The kinetic energy, \( K \), can then be calculated using the formula \( K =\frac{1}{2}I\omega^2 \).

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Key Concepts

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

Rotational Motion
Rotational motion refers to the movement of an object around a central point or pivot. This movement is best described by a set of rotational quantities similar to linear motion counterparts, but that are specific to rotation. Rather than covering a distance in meters, objects in rotational motion cover an angle in radians.

Some key parameters involved in rotational motion include:
  • Angular displacement (\(\alpha\)): The measure of the angle through which an object has turned, often measured in radians.
  • Angular velocity (\(\omega\)): How fast an object rotates, measured in radians per second.
  • Angular acceleration (\(a\)): The rate of change of angular velocity, similar to acceleration in linear motion, but for rotation.
These parameters are used in equations much like those in linear motion. For example, the equation \( \omega^2 = \omega_0^2 + 2\cdot \alpha\cdot a \) is a rotational version of the linear motion equation \( v^2 = u^2 + 2as \). Understanding these parallels can help in grasping concepts of rotational dynamics.
Moment of Inertia
Moment of inertia is a critical concept in rotational dynamics and can be thought of as the rotational equivalent of mass in linear motion. It reflects how the mass of an object is distributed with respect to the axis of rotation, determining how much torque is needed for a given angular acceleration.

For different shapes, there are specific formulas to calculate the moment of inertia. For example:
  • Solid sphere: \( I = \frac{2}{5}mr^2 \)
  • Hollow sphere: \( I = \frac{2}{3}mr^2 \)
  • Cylinder: \( I = \frac{1}{2}mr^2 \)
In our problem, the hollow sphere has its moment of inertia calculated using the formula \( I = \frac{2}{3}mr^2 \). This reflects how its mass is distributed towards the outer part of the sphere rather than evenly throughout like a solid sphere. Grasping this concept helps one appreciate how different objects react to forces when they are set into rotational motion.
Angular Velocity
Angular velocity describes how fast an object is rotating. It is a key component in understanding rotational motion, as it directly relates to the rotational kinetic energy of an object. The symbol \( \omega \) is often used to represent angular velocity, and it is typically measured in radians per second (rad/s).

To determine angular velocity in practical scenarios, particularly when starting from rest or under constant acceleration, the relationship \( \omega^2 = \omega_0^2 + 2\cdot \alpha\cdot a \) is used:
  • \( \omega_0 \) represents the initial angular velocity, which is zero if beginning from rest.
  • \( \alpha \) indicates the total angular displacement.
  • \( a \) denotes the angular acceleration.
In the given exercise, the shell begins at rest, so \( \omega_0 = 0 \). By using other known values in the formula, we can find its final angular velocity. This velocity is crucial for calculating the rotational kinetic energy, which in turn tells us how energy is stored as the object moves.

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Most popular questions from this chapter

Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits \(10^{-7} \mathrm{~m}\) deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are \(25.0 \mathrm{~mm}\) and \(58.0 \mathrm{~mm}\), respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of \(1.25 \mathrm{~m} / \mathrm{s}\). (a) What is the angular speed of the \(\mathrm{CD}\) when the innermost part of the track is scanned? The outermost part of the track? (b) The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration \(\mathrm{CD}\) if it were stretched out in a straight line? (c) What is the average angular acceleration of a maximum-duration CD during its 74.0 min playing time? Take the direction of rotation of the disc to be positive.

An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev \(/\) min to 200 rev \(/ \min\) in 4.00 s. (a) Find the angular acceleration in rev/s \(^{2}\) and the number of revolutions made by the motor in the 4.00 s interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?

A meter stick with a mass of \(0.180 \mathrm{~kg}\) is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released. As it swings through the vertical, calculate (a) the change in gravitational potential energy that has occurred; (b) the angular speed of the stick; (c) the linear speed of the end of the stick opposite the axis. (d) Compare the answer in part (c) to the speed of a particle that has fallen \(1.00 \mathrm{~m}\), starting from rest.

A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass \(40.0 \mathrm{~kg}\) and diameter \(75.0 \mathrm{~cm}\). The power is off for \(30.0 \mathrm{~s}\), and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 200 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on? (b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Measuring \(I\). As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be \(0.640 \mathrm{~m}\). Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an \(8.20 \mathrm{~kg}\) block of wood from the free end of the rope, as in Fig. E9.49. You release the system from rest and find that the block descends \(12.0 \mathrm{~m}\) in \(4.00 \mathrm{~s}\). What is the moment of inertia of the wheel for this axis?

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