Problem 97 - University Physics with Modern Physics

Question

CALC A slender, uni- form, metal rod with mass \(M\) is pivoted without friction about an axis through its midpoint and perpendicular to the rod. A horizontal spring with force constant \(k\) is attached to the lower end of the rod, with the other end of the spring attached to a rigid support. If the rod is displaced by a small angle \(\theta\) from the vertical (Fig. \(P 14.97\) ) and released, show that it moves in angular SHM and calculate the period. (Hint: Assume that the angle \(\theta\) is small enough for the approximations \(\sin \theta \approx \theta\) and \(\cos \theta \approx 1\) to be valid. The motion is simple harmonic if \(d^{2} \theta / d t^{2}=-\omega^{2} \theta,\) and the period is then \(T=2 \pi / \omega .\) )

Step-by-Step Solution

Answer

The period of angular SHM for the rod is \( T = 2\pi \sqrt{\frac{M}{12k}}.\)

Step 1: Determine the Restoring Torque

When the rod is displaced by an angle \( \theta \), the spring generates a restoring force \( F = -k x \), where \( x \approx L \theta \) for small angles. The torque due to this force is \( \tau = -k L \theta \cdot L = -k L^2 \theta \), acting to return the rod to equilibrium.

Step 2: Apply the Rotational Form of Newton's Second Law

Using Newton's second law for rotation, we have \( I \frac{d^2 \theta}{dt^2} = \tau \). The moment of inertia \( I \) for a rod pivoted about its midpoint is \( \frac{1}{12} ML^2 \). Thus, \( \frac{1}{12} ML^2 \frac{d^2 \theta}{dt^2} = -k L^2 \theta \).

Step 3: Simplify the Angular Equation

Divide through by \( L^2 \) to simplify: \( \frac{1}{12} M \frac{d^2 \theta}{dt^2} = -k \theta \). Rearrange to \( \frac{d^2 \theta}{dt^2} = -\frac{12k}{M} \theta \).

Step 4: Identify the Angular Frequency

Recognize the equation \( \frac{d^2 \theta}{dt^2} = -\omega^2 \theta \) for simple harmonic motion and compare it to \( \frac{d^2 \theta}{dt^2} = -\frac{12k}{M} \theta \). Thus, \( \omega^2 = \frac{12k}{M} \) and \( \omega = \sqrt{\frac{12k}{M}} \).

Step 5: Calculate the Period of SHM

The period \( T \) is given by \( T = \frac{2\pi}{\omega} \). Substituting for \( \omega \), \[ T = \frac{2\pi}{\sqrt{\frac{12k}{M}}} = 2\pi \sqrt{\frac{M}{12k}}. \]

Key Concepts

Torque

Torque is a pivotal concept in rotational dynamics. It refers to the rotational version of force, essentially describing how much a force causes an object to rotate. When you try to open a door, you apply a force at a distance from the hinge. That action creates torque, which turns the door. Mathematically, torque (\( \tau \)) is expressed as the product of the force (\( F \)) and the lever arm distance (\( r \)), the formula being \( \tau = r \cdot F \cdot \sin(\theta) \). However, in angular simple harmonic motion, for simplifications, if the angle \( \theta \) is small enough, \( \sin(\theta) \approx \theta \), simplifying our calculations.
In this particular problem, the torque is crucial because it determines how the rod swings back to its equilibrium position. The spring attached to the rod applies a restoring force that acts through a distance equal to the length of the rod, creating a torque that opposes the displacement and pulls the rod back to its original position.

Moment of Inertia

The moment of inertia (\( I \)) is another cornerstone of rotational dynamics, parallel to mass in linear motion. It measures an object's resistance to changes in its rotational state, i.e., how hard it is to start or stop it spinning. Just as heavier objects are harder to accelerate in a straight line, objects with larger moments of inertia are harder to spin or stop spinning.
For the metal rod pivoted at its center, the moment of inertia is calculated using the formula \( I = \frac{1}{12} ML^2 \). This formula comes from integrating the mass distribution of the rod around its pivot point, highlighting how mass further from the axis contributes more to the moment of inertia.

Restoring Force

Restoring force is the force that acts to bring a system back to its equilibrium position. In the context of this exercise, the spring attached to the rod produces a restoring force according to Hooke's Law: \( F = -kx \). Here, \( k \) is the spring constant and \( x \) is the displacement from equilibrium, which in this case is the arc length related to the angle of displacement \( \theta \).
For small angles, \( x \approx L \theta \), simplifying the relation to \( F = -k L \theta \). This force creates a torque that acts on the rod, pulling it back toward its vertical starting position. The negative sign indicates that the force opposes displacement.

Rotational Dynamics

Rotational dynamics encompasses the principles and equations that govern the motion of rotating bodies. In essence, it is the rotational equivalent of linear dynamics. It involves applying Newton's laws to rotating systems. A key equation in rotational dynamics is \( I \frac{d^2 \theta}{dt^2} = \tau \), where \( I \) is the moment of inertia, \( \frac{d^2 \theta}{dt^2} \) is angular acceleration, and \( \tau \) is torque.
In our exercise, we used this equation to connect the moment of inertia and the restoring torque exerted by the spring. By solving the differential equation derived from this fundamental law, we determine how the rod behaves when displaced. Specifically, it shows the rod undergoes angular simple harmonic motion, meaning it swings back and forth with a regular period dependent on the rod's mass and the spring constant.