Problem 59 - University Physics with Modern Physics

Question

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

Step-by-Step Solution

Answer

(a) \( \frac{1}{12} ML^2 \); (b) \( \frac{7}{48} ML^2 \)."

Step 1: Understand the Physical Configuration

The rod is bent at its center into an "L" shape. So, it consists of two perpendicular segments, each of length \( \frac{L}{2} \), connected at the origin. We need to calculate the moment of inertia for different axes.

Step 2: Moment of Inertia through the Connection Point

For the axis passing through the point where the two segments meet (origin), we treat each segment separately. The moment of inertia of a straight rod of length \(L\) and mass \(M\) about an axis through one end perpendicular to the length is \( \frac{1}{3}ML^2 \).For half-length \( \frac{L}{2} \) and mass \( \frac{M}{2} \):\[ I_{segment} = \frac{1}{3} \left( \frac{M}{2} \right) \left( \frac{L}{2} \right)^2 = \frac{1}{24} ML^2 \].There are two such segments:\[ I_{total} = 2 \times \frac{1}{24} ML^2 = \frac{1}{12} ML^2 \].

Step 3: Moment of Inertia through the Midpoint of Connecting Line

To find the moment of inertia about the midpoint of the line connecting the two ends, let's locate this point. Since each segment is of length \( \frac{L}{2} \), the endpoints are at \( (\frac{L}{2},0) \) and \( (0,\frac{L}{2}) \). The midpoint is \( (\frac{L}{4}, \frac{L}{4}) \).Using the parallel axis theorem, the increase in moment of inertia due to moving the axis from the origin to this point is \(m(d^2)\) for each segment, where \( d \) is distance from segment center to the axis:\[ I_{parallel} = I_{segment} + \frac{M}{2} \left(\frac{L}{4}\right)^2 = \frac{1}{24} ML^2 + \frac{M}{2} \frac{L^2}{16} = \frac{1}{24} ML^2 + \frac{1}{32} ML^2 = \frac{7}{96} ML^2 \].The total moment of inertia with parallel axis contribution:\[ I_{total} = 2 \times \frac{7}{96} ML^2 = \frac{14}{96} ML^2 = \frac{7}{48} ML^2 \].

Key Concepts

Parallel Axis Theorem: Understand the Basics

The parallel axis theorem is a useful tool in physics for calculating the moment of inertia of an object about a new axis, given its moment of inertia about a parallel axis that passes through its center of mass. The theorem states:

• If you have an axis passing through the center of mass, the moment of inertia about any other parallel axis is the original moment of inertia plus the mass times the square of the distance between the two axes.
• Mathematically, it is expressed as: \[ I = I_{cm} + MD^2 \]where \( I \) is the moment of inertia about the new axis, \( I_{cm} \) is the moment of inertia about the axis through the center of mass, \( M \) is the mass, and \( D \) is the distance between the axes.

Understanding this theorem allows us to solve problems involving objects where the axis of rotation is not at the center of mass.
By applying this principle, you can easily find the moment of inertia at different axes based on their relative position to the original axis.

Exploring the Uniform Rod: Geometry and Mass Distribution

A uniform rod refers to a line segment where mass and geometrical properties are consistent throughout its length. In physics problems, this uniformity simplifies many calculations, especially when dealing with moments of inertia.

• A uniform rod means each part of the rod, no matter where, has equal mass per unit length, known as mass density.
• When a rod is bent, such as in the problem, into an "L" shape, each segment of the rod still maintains this uniform distribution, but now with different orientation.

For calculations, the mass of each segment must be taken into account separately: when the rod is bent at its center, each segment becomes half the original length, with half the total mass. This way of thinking helps in applying the moment of inertia formulas accurately for each segment.

Dealing with Perpendicular Segments in Physics Problems

When dealing with perpendicular segments, especially in the context of uniformly distributed mass like a rod, certain characteristics simplify calculating physical properties like moment of inertia.

• In the given problem, each segment of the rod forms an angle of 90 degrees with the other, essentially creating an "L" shape.
• By treating each segment independently, you can calculate the moment of inertia for each part and then sum them to find the total moment of inertia for the system.

Perpendicular segments create unique challenges but knowing how to handle each part individually simplifies finding solutions. Being comfortable with using geometrical properties, like the distance from a chosen axis, assists with accurate calculations.

Cracking an Educational Physics Problem

Solving educational physics problems often requires applying fundamental principles to new situations. The problem of the uniform rod bent into an "L" shape is a classic example of how different concepts intertwine.

• First, visualize and understand the physical setup. Knowing that each segment is treated as part of a whole is crucial.
• Second, apply known formulas and theorems, like the parallel axis theorem, in stages to build the solution.
• Lastly, double-check calculations against the problem's requirements to ensure your solution aligns with the setup.

Educational physics problems help reinforce core concepts while encouraging critical thinking and systematic problem-solving. Each problem tackled adds another layer of understanding to your physics knowledge.