Problem 55 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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: \[I_h = \frac{2}{3}MR^2\]

Step 3: 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 \( \frac{2}{3}MR^2 \).

Step 4: 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\[\frac{2}{3}MR^2 = \frac{2}{5}MR^2 + Md^2\]

Step 5: 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, \( d = \sqrt{\frac{4}{15}} R \).

Step 6: 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.

Key Concepts

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.