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Chapter 10: Problem 5

A \(500-\mathrm{g}\) uniform sphere of \(7.0-\mathrm{cm}\) radius spins frictionlessly at \(30 \mathrm{rev} / \mathrm{s}\) 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 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 \(7.0 \mathrm{~cm}\).

Short Answer

Expert verified
\(\text{KE}_r = 0.017 \, \text{kJ}, \ L = 0.18 \, \text{kg} \cdot \text{m}^2/s, \ k = 4.4 \, \text{cm}\)

Step by step solution

01

Calculate Moment of Inertia

For a uniform sphere with mass \( M = 0.5 \) kg and radius \( r = 0.07 \) m, the moment of inertia \( I \) about its center axis is given as \[ I = \frac{2}{5} M r^2 = (0.40)(0.50 \, \text{kg})(0.070 \, \text{m})^2 = 0.00098 \, \text{kg} \cdot \text{m}^2 \]
02

Angular Velocity Conversion

Convert the angular velocity from revolutions per second to radians per second: \[ \omega = 30 \, \text{rev/s} = 30 \times 2\pi \, \text{rad/s} = 188 \, \text{rad/s} \]
03

Kinetic Energy Calculation

Calculate the rotational kinetic energy using the formula \[ \text{KE}_r = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.00098 \, \text{kg} \cdot \text{m}^2 \times (188 \, \text{rad/s})^2 = 0.017 \, \text{kJ} \]
04

Angular Momentum Calculation

Calculate the angular momentum using \[ L = I \omega = 0.00098 \, \text{kg} \cdot \text{m}^2 \times 188 \, \text{rad/s} = 0.18 \, \text{kg} \cdot \text{m}^2/s \]
05

Radius of Gyration Calculation

Calculate the radius of gyration \( k \) using the formula \[ k = \sqrt{\frac{I}{M}} = \sqrt{\frac{0.00098 \, \text{kg} \cdot \text{m}^2}{0.50 \, \text{kg}}} = 0.044 \, \text{m} = 4.4 \, \text{cm} \]

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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 fundamental concept in rotational dynamics, acting as a rotational analogue to mass. It is a measure of an object's resistance to changes in its rotational state, similar to how mass measures resistance to changes in linear motion. For a given object, the moment of inertia depends on both the mass distribution and the axis about which the rotation occurs.
In the case of a uniform sphere, such as the one in the exercise, the formula to compute the moment of inertia about an axis through its center is given by \[I = \frac{2}{5} M r^2\]where \(M\) is the mass and \(r\) is the radius of the sphere. This formula shows that the moment of inertia is influenced by both the mass and the square of the radius, meaning that even small increases in the radius can lead to significant increases in rotational inertia.
Understanding the moment of inertia is crucial because it helps predict how easily different objects will rotate in response to torque, which is pivotal in analyzing any rotational system.
Angular Momentum
Angular momentum is a conserved quantity in physics, traditionally symbolized as \(L\), representing the momentum of an object in rotational motion. It combines both the rotational speed and the distribution of mass around the rotation axis, hence serving as a rotational counterpart to linear momentum.
The formula to determine angular momentum is given by \[L = I \omega\]where \(I\) is the moment of inertia and \(\omega\) is the angular velocity in radians per second. This highlights the direct relationship between angular momentum and rotational characteristics of an object.
In the given exercise, the sphere鈥檚 angular momentum is calculated using its moment of inertia and its angular velocity converted from revolutions per second to radians per second. Recognizing angular momentum is vital, especially in closed systems where it remains constant unless acted upon by an external torque. This conservation law is widely used in analyzing systems from microscopic particles to astronomical objects.
Radius of Gyration
The radius of gyration \(k\) is an intuitive parameter that relates to the distribution of an object's mass around an axis of rotation. It represents an equivalent radius where the entire mass of the body could be concentrated to produce the same moment of inertia.To find the radius of gyration, we use the formula:\[k = \sqrt{\frac{I}{M}}\]where \(I\) is the moment of inertia and \(M\) is the mass of the object. This expression provides a single value that simplifies understanding how the mass is distributed relative to the rotation axis.
In our exercise, we see that the radius of gyration for the given sphere is smaller than its actual radius. This hints at how the mass is predominantly concentrated towards the center rather than the outer surface, common for solid objects like spheres. Grasping the concept of the radius of gyration is particularly useful in structural engineering and biomechanics, among other fields, where understanding mass distribution is key to stability and performance.

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Most popular questions from this chapter

As shown in Fig. \(10-3\), a mass \(m=400 \mathrm{~g}\) hangs from the rim of a frictionless pulley of radius \(r=15 \mathrm{~cm}\). When released from rest, the mass falls \(2.0 \mathrm{~m}\) in \(6.5 \mathrm{~s}\). Find the moment of inertia of the wheel. The hanging mass linearly accelerates downward due to its weight, and the pulley angularly accelerates clockwise due to the torque produced by the rope). The two motions are linked by the fact that \(a_{T}=r \alpha\). Consequently we will need to determine \(a_{T}\), and then \(\alpha\), and then \(F_{T}\), and then \(\tau\), and then \(I\). Remember that Newton's Second Law is central here (i.e., \(\tau=I \alpha\) for the wheel and \(F=m a\) for the mass). First we find \(a\) using \(y=v_{i} t+\frac{1}{2} a t^{2}\), since the mass accelerates down uniformly: $$ 2.0 \mathrm{~m}=0+\frac{1}{2} a(6.5 \mathrm{~s})^{2} $$ which yields \(a=0.095 \mathrm{~m} / \mathrm{s}^{2}\), and that equals the tangential acceleration \(\left(a_{T}\right)\) of a point on the rim of the pulley, which equals the acceleration \(a\) of the rope. Then, from \(a_{T}=\alpha r\), $$ \alpha=\frac{a_{T}}{r}=\frac{0.095 \mathrm{~m} / \mathrm{s}^{2}}{0.15 \mathrm{~m}}=0.63 \mathrm{rad} / \mathrm{s}^{2} $$ The net force on the mass \(m\) is \(m g-F_{T}\) and so \(F=m a\) becomes $$ \begin{aligned} m g-F_{T} &=m a_{T} \\ (0.40 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)-F_{T} &=(0.40 \mathrm{~kg})\left(0.095 \mathrm{~m} / \mathrm{s}^{2}\right) \end{aligned} $$ from which it follows that \(F_{T}=3.88 \mathrm{~N}\). Now \(\tau=I \alpha\) for the wheel: $$ \left(F_{T}\right)(r)=I \alpha \quad \text { or } \quad(3.88 \mathrm{~N})(0.15 \mathrm{~m})=I\left(0.63 \mathrm{rad} / \mathrm{s}^{2}\right) $$ from which we get \(I=0.92 \mathrm{~kg} \cdot \mathrm{m}^{2}\)

A thin cylindrical hoop having a diameter of \(1.0 \mathrm{~m}\) and a mass of \(400 \mathrm{~g}\), rolls down the street. What is the hoop's moment of inertia about its central axis of rotation? It follows from the first part of Fig. \(10-1\) that for a hoop $$ I=M R^{2}=(0.400 \mathrm{~kg})(0.50 \mathrm{~m})^{2}=0.10 \mathrm{~kg} \cdot \mathrm{m}^{2} $$

The driving side of a belt has a tension of \(1600 \mathrm{~N}\), and the slack side has 500 - \(\mathrm{N}\) tension. The belt turns a pulley \(40 \mathrm{~cm}\) in radius at a rate of \(300 \mathrm{rpm} .\) This pulley drives a dynamo having 90 percent efficiency. How many kilowatts are being delivered by the dynamo?

As a solid disk rolls up and over the top of a hill on a track, its speed slows to \(80 \mathrm{~cm} / \mathrm{s}\). It subsequently descends down the other side of the hill. If friction losses are negligible, how fast is the disk moving when it is \(18 \mathrm{~cm}\) below the top? At the top, the disk has translational and rotational \(\mathrm{KE}\), plus its \(\mathrm{PE}_{\mathrm{G}}\) relative to the point \(18 \mathrm{~cm}\) below. At that final point, \(\mathrm{PE}_{\mathrm{G}}\) has been transformed to more \(\mathrm{KE}\) of rotation and translation. Conservation of energy can be expressed as $$ \begin{array}{l} \left(\mathrm{KE}_{t}+\mathrm{KE}_{r}\right)_{\text {start }}+M g h=\left(\mathrm{KE}_{t}+\mathrm{KE}_{r}\right)_{\text {end }} \\ \frac{1}{2} M v_{i}^{2}+\frac{1}{2} I \omega_{i}^{2}+M g h=\frac{1}{2} M v_{f}^{2}+\frac{1}{2} I \omega_{f}^{2} \end{array} $$ For a solid disk, \(I=\frac{1}{2} M r^{2}\). Also, \(\omega=v / r\). Substituting these values and simplifying yields $$ \frac{1}{2} v_{i}^{2}+\frac{1}{4} v_{i}^{2}+g h=\frac{1}{2} v_{f}^{2}+\frac{1}{4} v_{f}^{2} $$ Employing \(v_{i}=0.80 \mathrm{~m} / \mathrm{s}\) and \(h=0.18 \mathrm{~m}\), substitution gives \(v_{f}=1.7 \mathrm{~m} / \mathrm{s}\).

An airplane propeller has a mass of \(70 \mathrm{~kg}\) and a radius of gyration of \(75 \mathrm{~cm}\). Find its moment of inertia. How large a torque is needed to give it an angular acceleration of \(4.0 \mathrm{rev} / \mathrm{s}^{2} ?\) $$ I=M k^{2}=(70 \mathrm{~kg})(0.75 \mathrm{~m})^{2}=39 \mathrm{~kg} \cdot \mathrm{m}^{2} $$ To be able to use \(\tau=I \alpha\), we must have \(\alpha\) in \(\mathrm{rad} / \mathrm{s}^{2}\) : $$ \alpha=\left(4.0 \frac{\mathrm{rev}}{\mathrm{s}^{2}}\right)\left(2 \pi \frac{\mathrm{rad}}{\mathrm{rev}}\right)=8.0 \pi \mathrm{rad} / \mathrm{s}^{2} $$ Then \(\tau=I \alpha=\left(39 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left(8.0 \pi \mathrm{rad} / \mathrm{s}^{2}\right)=0.99 \mathrm{kN} \cdot \mathrm{m}\)
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