91Ó°ÊÓ

A 500 -g uniform sphere of \(7.0\) -cm radius spins frictionlessly at 30 rev/s on an axis through its center. Find its ( \(a\) ) \(\mathrm{KE}_{r}\), (b) angular momentum, and ( \(c\) ) radius of gyration. We need the moment of inertia of a uniform sphere about an axis through its center. From \(\underline{\text { Fig. } 10-1}\), $$I=\frac{2}{5} M r^{2}=(0.40)(0.50 \mathrm{~kg})(0.070 \mathrm{~m})^{2}=0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}$$ (a) Knowing that \(\omega=30 \mathrm{rev} / \mathrm{s}=188 \mathrm{rad} / \mathrm{s}\), we have $$\mathrm{KE}_{r}=\frac{1}{2} I \omega^{2}=\frac{1}{2}\left(0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(188 \mathrm{rad} / \mathrm{s})^{2}=0.017 \mathrm{~kJ}$$ Notice that \(\omega\) must be in \(\mathrm{rad} / \mathrm{s}\). (b) Its angular momentum is $$L=I \omega=\left(0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(188 \mathrm{rad} / \mathrm{s})=0.18 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}$$ (c) For any object, \(I=M k^{2}\), where \(k\) is the radius of gyration. Therefore, $$k=\sqrt{\frac{I}{M}}=\sqrt{\frac{0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}}{0.50 \mathrm{~kg}}}=0.044 \mathrm{~m}=4.4 \mathrm{~cm}$$ Notice that this is a reasonable value in view of the fact that the radius of the sphere is \(7.0 \mathrm{~cm}\).

Step by step solution

01

Determine Moment of Inertia

The moment of inertia of a uniform sphere about an axis through its center is given by the formula \( I = \frac{2}{5} M r^{2} \). We have\[ M = 0.50 \text{ kg} \quad \text{and} \quad r = 0.070 \text{ m} \]Calculate \( I \) using the provided values:\[ I = \frac{2}{5} \times 0.50 \times (0.070)^{2} = 0.00098 \text{ kg} \cdot \text{m}^{2} \]

02

Calculate Angular Velocity

The angular velocity in radians per second is \( \omega = 30 \text{ rev/s} \). Convert this to radians per second:\[ \omega = 30 \times 2\pi = 188 \text{ rad/s} \]

03

Compute Rotational Kinetic Energy

The rotational kinetic energy is given by:\[ \mathrm{KE}_{r} = \frac{1}{2} I \omega^{2} \]Substitute the values for \( I \) and \( \omega \):\[ \mathrm{KE}_{r} = \frac{1}{2} \times 0.00098 \times (188)^{2} \approx 0.017 \text{ kJ} \]

04

Calculate Angular Momentum

Angular momentum \( L \) is calculated using:\[ L = I \omega \]Plug in the known values:\[ L = 0.00098 \times 188 = 0.18 \text{ kg} \cdot \text{m}^{2} / \text{s} \]

05

Determine Radius of Gyration

The radius of gyration \( k \) is found by equating moment of inertia to \( M k^{2} \):\[ k = \sqrt{\frac{I}{M}} \]Using the given \( I \) and \( M \):\[ k = \sqrt{\frac{0.00098}{0.50}} = 0.044 \text{ m} \approx 4.4 \text{ cm} \]

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Key Concepts

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

Moment of Inertia

The moment of inertia is a key concept in rotational dynamics. It's essentially the rotational equivalent of mass for linear motion. This concept describes how much torque is needed for a certain angular acceleration of an object around a particular axis. For a uniform sphere, the formula to calculate the moment of inertia about an axis through its center is:\[ I = \frac{2}{5} M r^{2} \]Here, \(M\) is the mass, and \(r\) is the radius of the sphere. Understanding moment of inertia helps in predicting how difficult it is to start spinning, stop, or change the speed of rotation for an object. In this exercise, we see how to compute it by substituting the sphere's mass and radius into the formula, leading to \(I = 0.00098 \text{ kg}\cdot\text{m}^{2}\). This tells us how the mass is distributed in relation to the axis of rotation.

Angular Momentum

Angular momentum is a crucial concept in rotational motion, akin to linear momentum in straight-line movement. It measures the quantity of rotation an object has, considering its moment of inertia and angular velocity. Mathematically, it's defined as:\[ L = I \omega \]where \(I\) is the moment of inertia, and \(\omega\) is the angular velocity. In practical terms, angular momentum helps us understand how the size and speed of a rotating object affect its rotation. A higher moment of inertia or faster rotation results in greater angular momentum. In our sphere example, with an angular velocity of \(188 \text{ rad/s}\), the angular momentum calculated as \(0.18 \text{ kg}\cdot\text{m}^{2}/\text{s}\) illustrates the object's capacity to keep spinning. Angular momentum is conserved in a closed system, making it a fundamental principle in physics to analyze rotational dynamics.

Kinetic Energy

Kinetic energy in the context of rotation is known as rotational kinetic energy. It's the energy due to the rotation of an object and depends on the moment of inertia and the square of angular velocity. The equation is:\[ \mathrm{KE}_{r} = \frac{1}{2} I \omega^{2} \]where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.This form of energy represents how movement is stored in a rotating body. In our example with the spinning sphere, substituting the moment of inertia and angular velocity, we find the rotational kinetic energy as approximately \(0.017 \text{ kJ}\).This energy plays a vital role in systems involving rotating parts, like engines and turbines, and understanding it allows the calculation of how much work can be done by these rotating systems.

Radius of Gyration

The radius of gyration is a measure that helps to simplify the analysis of rotational motion. It denotes a hypothetical distance from the axis of rotation at which the entire mass of an object could be assumed to be concentrated, without changing its moment of inertia. The formula for the radius of gyration \(k\) is:\[ k = \sqrt{\frac{I}{M}} \]where \(I\) is the moment of inertia, and \(M\) is the mass of the object.This measure is particularly useful for comparing how different shapes distribute mass relative to a central axis. In the case of our sphere, with a calculated radius of gyration of \(4.4 \text{ cm}\), it indicates how closely the mass distribution approximates a concentrated mass at that distance. Understanding the radius of gyration is helpful in engineering and physics for analyzing stability and stress distribution in rotating systems.