Problem 93 - University Physics with Modern Physics

Question

A thin, flat, uniform disk has mass \(M\) and radius \(R\) . A circular hole of radius \(R / 4,\) centered at a point \(R / 2\) from the disk's center, is then punched in the disk. (a) Find the moment of inertia of the disk with the hole about an axis through the original center of the disk, perpendicular to the plane of the disk. (Hint: Find the moment of inertia of the piece punched from the disk. (b) Find the moment of inertia of the disk with the hole about an axis through the center of the hole, perpendicular to the plane of the disk.

Step-by-Step Solution

Answer

(a) \( \frac{247}{512} MR^2 \); (b) \( \frac{487}{512} MR^2 \).

Step 1: Moment of Inertia of the Full Disk

The moment of inertia of a full disk of mass \( M \) and radius \( R \) about an axis through the center, perpendicular to its plane, is given by \( I_\text{full} = \frac{1}{2} M R^2 \).

Step 2: Moment of Inertia of the Punched Hole

The punched hole is a smaller disk of radius \( R/4 \). The mass of the punched hole can be calculated using the area ratio: \( M_\text{hole} = M \cdot \frac{(\pi (R/4)^2)}{\pi R^2} = M/16 \). The moment of inertia of a disk about an axis through its center is \( \frac{1}{2} M R^2 \). Therefore, the moment of inertia of the hole about its own center is \( I_\text{hole center} = \frac{1}{2} \cdot \frac{M}{16} \cdot (R/4)^2 = \frac{M R^2}{512} \).

Step 3: Apply Parallel Axis Theorem to Punched Hole

To find the moment of inertia of the hole about the original axis (the center of the large disk), we use the parallel axis theorem: \( I_\text{hole center} + M_\text{hole} \cdot d^2 \), where \( d = R/2 \) is the distance from the hole's center to the original disk's center. Thus, \( I_\text{hole original} = \frac{M R^2}{512} + \frac{M}{16} \cdot \left(\frac{R}{2}\right)^2 = \frac{M R^2}{512} + \frac{M R^2}{64} = \frac{9M R^2}{512} \).

Step 4: Moment of Inertia of Disk with Hole (a)

Subtract the moment of inertia of the punched-out hole from the moment of inertia of the full disk: \( I_\text{with hole} = I_\text{full} - I_\text{hole original} = \frac{1}{2} M R^2 - \frac{9M R^2}{512} = \frac{247MR^2}{512} \).

Step 5: Moment of Inertia of Disk with Hole about Hole's Center (b)

To find the moment of inertia about an axis through the center of the hole, shift the axis offset by using the parallel axis theorem again: \( I_\text{hole new} = I_\text{with hole} + M_\text{remaining} \cdot d^2 \), where \( M_\text{remaining} = M - \frac{M}{16} = \frac{15M}{16} \) and \( d = \frac{R}{2} \). Thus, \( I_\text{new} = \frac{247MR^2}{512} + \frac{15M}{16} \cdot \left(\frac{R}{2}\right)^2 = \frac{247MR^2}{512} + \frac{15MR^2}{64} = \frac{487MR^2}{512} \).

Key Concepts

Parallel Axis Theorem

The Parallel Axis Theorem is a key principle in rotational dynamics. It is incredibly helpful when calculating the moment of inertia for an object relative to an axis that is parallel to an axis through its center of mass.
This theorem states that the moment of inertia about any axis parallel to an axis through the center of mass is the sum of two components:

• The moment of inertia about the center of mass, denoted as \( I_{CM} \).
• The product of the mass \( m \) and the square of the distance \( d \) between these axes. This is expressed as \( m \, d^2 \).

The formula is given by:\[I = I_{CM} + m \, d^2\] In the problem, you use the Parallel Axis Theorem to determine the moment of inertia of the punched hole about the original center of the disk. By calculating \( I_{hole original} \) with this theorem, you find how the axis shift affects the inertia due to the hole's offset from the central axis of the original disk.

Rotational Dynamics

Rotational dynamics involves the motion of objects that rotate around an axis. Understanding the moment of inertia plays a crucial role as it affects how easily an object can rotate.
The moment of inertia is essentially a measure of an object's resistance to changes in its rotation. It depends on how the object's mass is distributed in relation to its rotation axis. In this exercise, you're dealing with a uniform disk from which a circular portion has been removed. To solve for the rotational dynamics of such a system, one needs to analyze how the subtraction of the mass changes the moment of inertia.

• The moment of inertia for a full disk is initially calculated about its central axis.
• Then, you find the inertia of the missing piece and subtract it from the full disk's inertia to account for the absent section.

This approach ensures an accurate representation of the disk's ability to resist rotational changes after the hole is punched out.

Mass and Radius Calculation

To solve for the moment of inertia involving shapes with holes or offsets, the mass and radius of these elements must be known. Calculations often require understanding the relationships between total mass and the mass of the portion removed or added.
For the disk and hole problem:

• First, you determine the mass of the full disk \( M \) and its radius \( R \).
• The mass of the punched hole is calculated using proportional areas. The smaller disk (the hole) has a radius of \( R/4 \), thus its mass \( M_{hole} \) is \( M/16 \).

With these values, you can proceed to calculate moments of inertia by knowing the dimensions and the mass distribution. This mathematical groundwork is crucial for applying the Parallel Axis Theorem and making further calculations for rotational problems. Mastery of this makes tackling problems of rotation involving varied materials and cross-sectional shapes much more straightforward.