Problem 53 - University Physics with Modern Physics

Question

A 1.80 -kg connecting rod from a car engine is pivoted about a horizontal knife edge as shown in Fig. E14.53. The center of gravity of the rod was located by balancing and is 0.200 \(\mathrm{m}\) from the pivot. When the rod is set into small-amplitude oscillation, it makes 100 complete swings in 120 s. Calculate the moment of inertia of the rod about the rotation axis through the pivot.

Step-by-Step Solution

Answer

The moment of inertia is approximately 0.1342 kg⋅m².

Step 1: Calculate the Period of Oscillation

First, determine the time period of one complete oscillation by dividing the total time by the number of swings. The rod makes 100 complete oscillations in 120 seconds, thus the period \( T \) is given by \[ T = \frac{120 \, \mathrm{s}}{100} = 1.2 \, \mathrm{s} \]

Step 2: Use the Formula for the Period of a Physical Pendulum

For a physical pendulum, the period of oscillation is given by the equation \( T = 2\pi \sqrt{\frac{I}{mgd}} \), where \( I \) is the moment of inertia, \( m \) is the mass of the rod, \( g \) is the acceleration due to gravity (\(9.81 \, \mathrm{m/s^2}\)), and \( d \) is the distance from the pivot to the center of gravity (\(0.200 \, \mathrm{m}\)).

Step 3: Rearrange the Formula to Solve for Moment of Inertia

Rearrange the equation to solve for the moment of inertia \( I \): \[ I = \frac{T^2 \cdot mgd}{4\pi^2} \]

Step 4: Substitute the Known Values into the Formula

Substitute the given values into the formula: \( T = 1.2 \, \mathrm{s} \), \( m = 1.80 \, \mathrm{kg} \), \( g = 9.81 \, \mathrm{m/s^2} \), and \( d = 0.200 \, \mathrm{m} \). Now calculate: \[ I = \frac{(1.2 \, \mathrm{s})^2 \times 1.80 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} \times 0.200 \, \mathrm{m}}{4\pi^2} \] \[ I \approx \frac{5.30032}{39.4784} \approx 0.1342 \, \mathrm{kg} \cdot \mathrm{m^2} \]

Step 5: Present the Final Result

The moment of inertia of the rod about the rotation axis through the pivot is approximately \( I = 0.1342 \, \mathrm{kg} \cdot \mathrm{m^2} \).

Key Concepts

Physical Pendulum

A physical pendulum is a type of pendulum where the mass is distributed along the entire length of the object rather than concentrated at a single point. This distinguishes it from a simple pendulum, which assumes all the mass is concentrated at its center of mass. In the exercise, the connecting rod from a car engine serves as the physical pendulum.

When such a pendulum swings, it moves back and forth around its pivot point. This motion is dictated by several factors, including the distribution of mass and the gravitational pull.

• The entire piece acts as it swings, with each part contributing to the overall moment of inertia.
• The formula for a physical pendulum includes parameters such as moment of inertia and the distance from the pivot to the center of gravity.

Understanding how these factors interplay is crucial for calculations involving the moment of inertia, which defines how much torque is needed for a desired angular acceleration.

Oscillation Period

The oscillation period is the time it takes for the pendulum to return to its initial position after one complete cycle of motion. In this scenario, the exercise mentions the rod making 100 complete swings in 120 seconds.

To find the period of a single oscillation, we divide the total time by the number of oscillations. Thus, the period here is 1.2 seconds.

• The oscillation period provides insights into the pendulum's rhythmic motion.
• For a physical pendulum, the period is described by the formula: \[ T = 2\pi \sqrt{\frac{I}{mgd}} \]
• This formula captures the relationship between the period and factors such as mass, gravitational force, and the distance from the pivot to the center of gravity.

Calculating the period is fundamental in analyzing and predicting the motion of the pendulum over time.

Center of Gravity

The center of gravity is a crucial point in an object where its entire weight is considered to be concentrated. For a physical pendulum, this point plays a pivotal role in understanding its oscillation.

In our exercise, the center of gravity of the car engine rod is located 0.200 m from the pivot. This distance is essential in the period calculation formula, since it determines the leverage effect of gravity on the pendulum's swing.

• The center of gravity can shift, affecting how the pendulum behaves.
• It is determined through balancing or calculation based on the object's mass distribution.
• For accurate results, it's important to recognize the placement of the center of gravity in relation to the pivot.

Insight into the center of gravity allows for the precise modeling of the pendulum’s motion and is key to solving related physics problems.