Problem 38 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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}. \]

Step 2: 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)}. \]

Step 3: 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}. \]

Step 4: 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. \]

Step 5: 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. \]

Step 6: 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}. \]

Key Concepts

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.