Explain the concept of compound pendulum, also write the procedure for determine radius of gyration about an axis through the center of gravity for the compound pendulum

 

Concept of Compound Pendulum

A compound pendulum, also known as a physical pendulum, is any rigid body that is free to oscillate about a horizontal axis that does not pass through its center of mass. Unlike a simple pendulum, which consists of a mass (bob) at the end of a weightless string, a compound pendulum takes into account the distribution of mass along the length of the pendulum.

Key characteristics of a compound pendulum include:

• Center of Mass (G): The point at which the total mass of the pendulum can be considered to be concentrated.
• Pivot Point (O): The fixed point around which the pendulum oscillates.
• Moment of Inertia (I): The measure of the resistance of the body to angular acceleration about the pivot point.
• Distance (h): The vertical distance between the pivot point and the center of mass.
• Period (T): The time it takes for the pendulum to complete one full oscillation.

The period of oscillation for a compound pendulum is given by the formula:

$T = 2\pi \sqrt{\frac{I}{mgh}}$

where:

• $T$ is the period of oscillation,
• $I$ is the moment of inertia about the pivot point,
• $m$ is the mass of the pendulum,
• $g$ is the acceleration due to gravity,
• $h$ is the distance between the pivot point and the center of mass.

Procedure to Determine the Radius of Gyration

The radius of gyration ($k$) of a compound pendulum about an axis through its center of gravity (G) is a measure of how the mass is distributed relative to the center of gravity. It can be determined through experimental measurements and calculations using the following steps:

1. Set Up the Pendulum:

   • Suspend the compound pendulum from a fixed pivot point so that it can oscillate freely.
   • Ensure that the pendulum can oscillate in a plane without any significant frictional forces acting on it.

2. Measure the Distance to the Center of Mass (h):

   • Determine the vertical distance $h$ between the pivot point and the center of mass of the pendulum. This can be done by balancing the pendulum on a knife edge and finding the point where it balances horizontally. The distance from this point to the pivot is $h$.

3. Measure the Period of Oscillation (T):

   • Displace the pendulum slightly from its equilibrium position and release it to start oscillating.
   • Use a stopwatch to measure the time it takes for the pendulum to complete several oscillations (e.g., 10 or 20 oscillations) to minimize the error.
   • Divide the total time by the number of oscillations to find the period $T$.

4. Calculate the Moment of Inertia (I):

   • Use the measured period $T$ and the known values of mass $m$, acceleration due to gravity $g$, and distance $h$ to calculate the moment of inertia about the pivot point using the formula:

   $I = \frac{T^2 mgh}{4\pi^2}$

5. Determine the Radius of Gyration (k):

   • The moment of inertia of the pendulum about an axis through its center of gravity can be related to the radius of gyration by the formula:

   $I_G = mk^2$

   • The parallel axis theorem relates the moment of inertia about the pivot point ($I$) to the moment of inertia about the center of gravity ($I_G$):

   $I = I_G + mh^2$

   • Substituting $I_G = mk^2$ into the equation gives:

   $I = mk^2 + mh^2$

   • Rearrange to solve for $k$:

   $k^2 = \frac{I}{m} - h^2$

   • Using the value of $I$ calculated earlier, determine the radius of gyration $k$:

   $k = \sqrt{\frac{I}{m} - h^2}$

Example Calculation

Suppose the following measurements were made:

• Mass of the pendulum, $m = 2 \, \text{kg}$
• Distance between the pivot point and the center of mass, $h = 0.5 \, \text{m}$
• Period of oscillation, $T = 2 \, \text{s}$
• Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

First, calculate the moment of inertia about the pivot point:

$I = \frac{T^2 mgh}{4\pi^2} = \frac{(2^2) (2)(9.81)(0.5)}{4\pi^2} = \frac{8 \times 9.81 \times 0.5}{39.4784} \approx 0.993 \, \text{kg}\cdot\text{m}^2$

Next, determine the radius of gyration:

$k^2 = \frac{I}{m} - h^2 = \frac{0.993}{2} - (0.5)^2 = 0.4965 - 0.25 = 0.2465$

$k = \sqrt{0.2465} \approx 0.496 \, \text{m}$

Thus, the radius of gyration $k$ about an axis through the center of gravity for the compound pendulum is approximately 0.496 meters.