Problem 52 - 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) \(\frac{ML^2}{12}\); (b) \(\frac{7ML^2}{12}\).

Step 1: Analyze the Rod Configuration

The rod is bent at its center into two segments, each of length \(\frac{L}{2}\), and these segments are perpendicular to each other forming an 'L' shape. The mass of each segment is \(\frac{M}{2}\) given the uniform distribution of mass.

Step 2: Moment of Inertia at the Corner

Consider the point where the two segments meet, creating a single axis perpendicular to the plane of the rod. Treat each segment separately and use the formula for the moment of inertia of a rod about its end point, \(I = \frac{1}{3}mL^2\). Here, \(m = \frac{M}{2}\) and \(L = \frac{L}{2}\). Add up the inertia contributions from both segments:\[I = 2 \times \left(\frac{1}{3} \times \frac{M}{2} \times \left(\frac{L}{2}\right)^2\right) = \frac{1}{3} \frac{ML^2}{4} = \frac{ML^2}{12}.\] Therefore, \(I_{corner} = \frac{ML^2}{12}.\)

Step 3: Moment of Inertia at the Midpoint of the Ends

Locate the midpoint of the line connecting the outer ends of the 'L' shape. The segments being perpendicular form a right-angled isosceles triangle with the diagonal from the origin of the corner to the center where the inertia is being considered. Thus, use the parallel axis theorem:\[I_{midpoint} = I_{corner} + M \times R^2,\] where \(R = \frac{L}{\sqrt{2}}.\) Therefore,\[I_{midpoint} = \frac{ML^2}{12} + M \left(\frac{L}{\sqrt{2}}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{2} = \frac{7ML^2}{12}.\]

Key Concepts

Parallel Axis Theorem

The parallel axis theorem is a critical tool in calculating moments of inertia of objects around axes not passing through the center of mass. It helps us find the moment of inertia about any axis parallel to an axis through the center of mass. The formula is given by:\[ I = I_{cm} + Md^2 \] where:

• \( I \) is the moment of inertia about the new axis.
• \( I_{cm} \) is the moment of inertia of the object around the center of mass axis.
• \( M \) is the total mass of the object.
• \( d \) is the perpendicular distance between the two axes.

In our problem, when considering the midpoint between the ends of the bent rod, the parallel axis theorem is applied. We first calculate the moment of inertia at the corner where the two segments meet and then account for the diagonal distance to the new axis. Understanding this shift in axis is key in many practical applications, such as engineering and physics problems.

Uniform Rod

A uniform rod refers to a rod with mass distributed evenly along its length. This characteristic simplifies calculations of physical properties, such as moments of inertia. For calculating moments of inertia of a uniform rod about different axes, we use various standard formulas.In this situation, the rod is bent into an 'L' shape, where each segment's mass and length are halved due to the uniform distribution. This condition is essential in calculating the individual contributions of inertia from each segment about a specific axis.For a straight uniform rod with length \( L \) and mass \( M \), the moment of inertia about an axis through one end perpendicular to its length is given by:\[ I = \frac{1}{3} ML^2 \]This formula is adapted for our case, where both segments are \( \frac{L}{2} \) in length and \( \frac{M}{2} \) in mass.

Perpendicular Axis

The configuration in our exercise requires considering two segments of a rod arranged perpendicularly. Each segment contributes individually to the overall moment of inertia about an axis that is also perpendicular to the plane formed by the segments. The perpendicular setup means we can't just treat the rod as one linear piece. Instead, we handle each segment separately. In calculation, this configuration allows us to treat moments about each segment's end and then consider them collectively. This setup is the basis of the concept called the "perpendicular axis theorem" which applies mainly in scenarios involving symmetry across two dimensions, which then integrates with the parallel axis theorem for 3D calculations.

Rotational Dynamics

Rotational dynamics explores the motion of bodies that rotate about an axis, providing insights into aspects like angular momentum, torque, and moment of inertia. The current exercise showcases how a modified shape of a uniform rod affects its rotational inertia. The bent rod exhibits dynamics different from a linear rod because of its shifted symmetry and two perpendicular sections. To predict how various forces influence its rotation, we calculate moments of inertia, considering the shape's complexity. Knowledge of this field helps solve problems related to machinery parts, leverage tools, and even celestial motion. For this rod, when evaluating rotational dynamics, dividing the rod into two parts provides a systematic approach. Each segment is assessed for how it contributes to the overall inertia. Such assessments lead to better understanding of stability and balance in rotating systems.