Question

A metal rod of length 'L' and mass ' \(\mathrm{m}\) ' is pivoted at one end. A thin disk of mass ' \(\mathrm{M}\) ' and radius ' \(R^{\prime}(\) Angular frequency for case \(B\). (D) Angular frequency for case \(A<\) Angular frequency for case \(B\).

Short answer

Statement (A) is true because the restoring torques are due to gravity acting on the mass centers which don't change. Statement (C) is true because case A has a larger moment of inertia, leading to a lower angular frequency than case B. Statements (B) and (D) are false.

Step-by-step solution

Introduction to the Problem

We are given a rod of length 'L' and mass 'm' that is pivoted at one end, with a disk of mass 'M' and radius 'R' attached to its free end. In case A, the disk isn't free to rotate, and in case B, it is. We need to compare the restoring torque and the angular frequency of the system for both cases performing Simple Harmonic Motion (SHM). We assume the system is only affected by gravity.

Define Restoring Torque and Angular Frequency

The restoring torque (\( \tau \)) in a rotational system performing SHM is given by \( \tau = -k\theta \), where \( k \) is the rotational equivalent of spring constant, and \( \theta \) is the angular displacement. The angular frequency (\( \omega \)) of SHM is given by \( \omega = \sqrt{\frac{k}{I}} \), where \( I \) is the moment of inertia of the system about the pivot point.

Restoring Torque for Both Cases

The gravitational restoring torque tends to return the system to the equilibrium position. Since gravity acts at the center of mass, and we're considering small-angle approximations, the restoring torques in both cases A and B are identical. This is because the mass distribution with respect to the pivot is unchanged in both cases; hence, the torque caused by the weight of the system would be the same.

Moment of Inertia for Case A

For case A, where the disk is not free to rotate, the moment of inertia \( I_A \) is the sum of the moment of inertia of the rod (pivot at the end) and the disk treated as rotating with the rod. It is given by \( I_A = \frac{1}{3}mL^2 + MR'^2 \).

Moment of Inertia for Case B

For case B, where the disk is free to rotate, the moment of inertia \( I_B \) is just the sum of the moment of inertia of the rod and the disk about its own center plus the additional distance from the pivot, which does not change because it's free to rotate. It is given by \( I_B = \frac{1}{3}mL^2 \).

Compare Angular Frequencies

Angular frequency (\( \omega \)) depends inversely on the square root of the moment of inertia. Since \( I_A > I_B \) (case A includes the inertia of the disk, whereas B does not), we can conclude that \( \omega_A < \omega_B \). Hence, the angular frequency for case A is less than that for case B.

Key concepts

Restoring Torque

In the context of rotational dynamics, the restoring torque plays a crucial role in Simple Harmonic Motion (SHM). It's the rotational counterpart to a restoring force in linear systems. When a system is displaced from its equilibrium position, restoring torque works to bring it back, much like a spring pulling an attached mass back to its rest position.

In the exercise, we consider the effect of gravity on a rod-disc assembly, which creates a restoring torque. The torque depends on the gravitational force acting at the system's center of mass and the distance from the pivot to this center of mass. For both case A and B, where a disk is attached at the end of a rod in different configurations, the center of mass and, accordingly, the restoring torque remains unchanged. This is because the gravitational force and lever arm are the same regardless of whether the disk is allowed to rotate freely or not.

Restoring Torque Consistency

Since the position of the mass components relative to the pivot remains identical in both cases, the restoring torque, caused by the weight of the system, does not change. This leads to the conclusion that the restoring torques for case A and B are equal, contrary to what might be assumed if one were not to consider the lever arm's constancy.

Moment of Inertia

The moment of inertia is a pivotal concept in rotational dynamics, representing an object's resistance to rotational acceleration about an axis. This is analogous to mass in linear motion; just as a larger mass requires more force to accelerate, a higher moment of inertia requires more torque to achieve the same angular acceleration.

In our problem, the difference between case A and B is all about how the moment of inertia is calculated. For case A, the disk does not rotate independently, so its mass contributes to the moment of inertia about the pivot. The formula for the moment of inertia of the rod-disc system in case A is a sum of the rod's inertia and the disk's inertia as if it was a part of the rod.

Free Versus Fixed Rotation

For case B, however, because the disk can rotate freely, its own moment of inertia doesn't add to that of the system’s rotation about the pivot. Therefore, the moment of inertia for case B is solely that of the rod, which is significantly less than in case A. This reduction in inertia is why the rod-disc assembly in case B would have a higher angular frequency compared to case A when performing SHM.

Simple Harmonic Motion

In physics, simple harmonic motion (SHM) is a type of periodic motion where the restoring force or torque is directly proportional to the displacement and acts in the direction opposite to that displacement. It's an idealization of the motion that occurs when an object is restored to equilibrium via a force or torque that is proportional to its displacement.

In the case of our rod-disc system, the SHM can be visualized when the assembly swings back and forth about the pivot point. Assuming small angular displacements, this motion can be described with the same mathematical equations used for linear SHM, with the understanding that forces and masses are replaced by torques and moments of inertia, respectively.

Angular SHM Characteristics

For both cases A and B, the motion is harmonic because the restoring torque remains proportional to the angular displacement. Yet, as we'll discuss next, the angular frequency of these motions differs due to the difference in the moments of inertia.

Rotational Dynamics

Now, let's tie it all together within the realm of rotational dynamics. This branch of mechanics addresses the behavior of objects as they rotate. The key variables are the angular velocity, the torque, and the moment of inertia, which parallels mass in linear motion equations.

The angular frequency, a fundamental quantity in SHM, describes how fast an object undergoes repetitive motion. As explored in the step-by-step solution, the angular frequency for case A is less than for case B. This conclusion comes from understanding that the moment of inertia for case A is greater due to the disk's fixed position relative to the rod, while in case B, the disk's rotation does not contribute to inertia of the system about the pivot.

Breaking Down Angular Frequency

Thus, a larger moment of inertia in case A results in a lower angular frequency compared to case B, according to the relationship \( \( \omega = \sqrt{\frac{k}{I}} \) \) where \(k\) is the rotational spring constant. It's an inverse relationship; as the moment of inertia increases, the angular frequency decreases, illustrating the property of rotational dynamics that heavier rotational masses (higher moments of inertia) rotate more slowly for a given restoring torque.