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Chapter 9: Problem 55

About what axis will a uniform, balsa-wood sphere have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius, with the axis along a diameter?

Short Answer

Expert verified
Shift the axis by \( d = \sqrt{\frac{4}{15}}R \) parallel to any diameter.

Step by step solution

01

Understanding the Problem

We are given two spheres: 1) a uniform balsa-wood sphere and 2) a thin-walled, hollow lead sphere. Both have the same mass \( M \) and radius \( R \). We need to determine the axis about which the uniform sphere will have the same moment of inertia as the hollow sphere does around its diameter.
02

Moment of Inertia for the Hollow Sphere

The moment of inertia \( I_h \) of a thin-walled, hollow sphere about an axis through its diameter is given by the formula: \[I_h = \frac{2}{3}MR^2\]
03

Moment of Inertia for the Solid Sphere

For the uniform solid sphere, the moment of inertia \( I_s \) about an axis through its center is:\[I_s = \frac{2}{5}MR^2\]We need to find the axis about which this sphere has a moment of inertia equal to \( \frac{2}{3}MR^2 \).
04

Parallel Axis Theorem Application

The parallel axis theorem states:\[ I = I_c + Md^2 \]where \( I_c \) is the moment of inertia through the center, \( M \) is the mass, and \( d \) is the perpendicular distance between axes. We apply this to find \( d \) such that\[\frac{2}{3}MR^2 = \frac{2}{5}MR^2 + Md^2\]
05

Solving for the Distance \(d\)

Subtract \( \frac{2}{5}MR^2 \) from both sides to get:\[Md^2 = \frac{2}{3}MR^2 - \frac{2}{5}MR^2 = \left(\frac{10}{15} - \frac{6}{15}\right)MR^2 = \frac{4}{15}MR^2\]Dividing by \( M \) and solving for \( d^2 \):\[d^2 = \frac{4}{15}R^2\]Thus, \( d = \sqrt{\frac{4}{15}} R \).
06

Finding the Axis

Since the moment of inertia is symmetric about any diameter, the axis should be shifted perpendicular to a diameter by \( d = \sqrt{\frac{4}{15}}R \) to have the same moment of inertia.

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Key Concepts

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

Parallel Axis Theorem
The Parallel Axis Theorem is an essential tool in rotational dynamics. This theorem allows us to calculate the moment of inertia of an object about any axis parallel to an axis through its center of mass (COM). The theorem states that:
  • If you know the moment of inertia of an object about its COM, labeled as \( I_c \), you can find its moment of inertia about a new axis (\( I \)) which is parallel and a distance \( d \) away, using the formula:\[ I = I_c + Md^2 \]
  • Here, \( M \) is the mass of the object, and \( d \) is the perpendicular distance between the two axes.
This theorem is particularly useful when dealing with composite bodies or when analyzing the rotation of bodies about an axis that is not directly through the center of mass. In our problem, it helps to determine how a solid sphere can have the same moment of inertia as a hollow sphere when the rotation axis is not through the center.
Solid Sphere
A solid sphere is a three-dimensional object where every point on its surface is equidistant from its center. This configuration affects its rotational dynamics. For a uniform solid sphere, the moment of inertia about a central axis is given by the formula:\[ I_s = \frac{2}{5}MR^2 \]
  • This formula reflects that a significant fraction of the mass is distributed closer to the center compared to hollow spheres.
  • Consequently, a solid sphere has a smaller moment of inertia compared to a hollow sphere of the same mass and radius when rotated about the same axis.
Understanding these distinctions is crucial when comparing rotational properties of different shapes, as done in the given exercise. The distribution of mass significantly influences how much torque is needed to achieve the same rotational effect.
Hollow Sphere
A hollow sphere, sometimes called a spherical shell, is essentially a sphere with no mass inside. All of its mass is concentrated on a thin shell at the outer boundary. The moment of inertia for a hollow sphere about an axis through its diameter is:\[ I_h = \frac{2}{3}MR^2 \]
  • This formula indicates that the mass distribution has a bigger radius of gyration compared to a solid sphere.
  • This difference in mass distribution results in the hollow sphere having a greater moment of inertia than a solid sphere of the same mass and radius.
In the context of our problem, understanding this principle is key in determining how to match the rotational characteristics of a solid sphere to those of a hollow one, by identifying the appropriate rotation axis.
Rotation Axis
The rotation axis is a fundamental concept when studying the moment of inertia and rotational dynamics. It is the line about which an object rotates. The object's moment of inertia depends significantly on where this axis is located relative to the mass distribution. Here are some key points about rotation axes:
  • For symmetrical objects like spheres, standard axes could be through the center or along any diameter.
  • In the exercise, we find an axis that makes a solid sphere equivalently resistant to rotation as a hollow sphere. To do this, we manipulated the axis position using the parallel axis theorem.
  • By calculating the required distance \( d \), we found that shifting the rotation axis perpendicularly by \( d = \sqrt{\frac{4}{15}}R \) achieves identical moment of inertia as that of a hollow sphere.
This adjustment shows how shifting the axis allows us to calibrate the rotational resistance or inertia to match that of a different configuration.

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Most popular questions from this chapter

The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208 \(\mathrm{m}\) is mounted on the same rotating shaft as the second pulley. (a) The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed? (b) Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't stick to its teeth.

A roller in a printing press turns through an angle \(\theta(t)\) given by \(\theta(t)=\gamma t^{2}-\beta t^{3},\) where \(\gamma=3.20 \operatorname{rad} / \mathrm{s}^{2}\) and \(\beta=\) 0.500 \(\mathrm{rad} / \mathrm{s}^{3}\) , (a) Calculate the angular velocity of the roller as a function of time. (b) Calculate the angular acceleration of the roller as a function of time. (c) What is the maximum positive angular velocity, and at what value of \(t\) does it occur?

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 CD when the innermost part of the track is scanned? The outermost part of the track? (b) The maximum playing time of a \(\mathrm{CD}\) 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 dise to be positive.

At \(t=0\) a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0 \(\mathrm{rad} / \mathrm{s}^{2}\) until a circuit breaker trips at \(t=2.00 \mathrm{s}\) . From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between \(t=0\) and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

A pendulum is made of a uniform solid sphere with mass \(M\) and radius \(R\) suspended from the end of a light rod. The distance from the pivot at the upper end of the rod to the center of the sphere is \(L\) . The pendulum's moment of inertia I for rotation about the pivot is usually approximated as \(M L^{2} .\) (a) Use the parallel-axis theorem to show that if \(R\) is 5\(\%\) of \(L\) and the mass of the rod is ignored, \(I_{p}\) is only 0.1\(\%\) greater than \(M L^{2} .\) (b) If the mass of the rod is 1\(\%\) of \(M\) and \(R\) is much less than \(L,\) what is the ratio of \(I_{\text { rod }}\) for an axis at the pivot to \(M L^{2} ?\)
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