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Chapter 14: Problem 53

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.

Short Answer

Expert verified
The moment of inertia is approximately 0.1342 kg鈰卪虏.

Step by step solution

01

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 \[ T = \frac{120 \, \mathrm{s}}{100} = 1.2 \, \mathrm{s} \]
02

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 (\(0.200 \, \mathrm{m}\)).
03

Rearrange the Formula to Solve for Moment of Inertia

Rearrange the equation to solve for the moment of inertia \( I \): \[ I = \frac{T^2 \cdot mgd}{4\pi^2} \]
04

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: \[ I = \frac{(1.2 \, \mathrm{s})^2 \times 1.80 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} \times 0.200 \, \mathrm{m}}{4\pi^2} \] \[ I \approx \frac{5.30032}{39.4784} \approx 0.1342 \, \mathrm{kg} \cdot \mathrm{m^2} \]
05

Present the Final Result

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 motion and is key to solving related physics problems.

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Most popular questions from this chapter

CP A rocket is accelerating upward at 4.00 \(\mathrm{m} / \mathrm{s}^{2}\) from the launchpad on the earth. Inside a small, 1.50 -kg ball hangs from the ceiling by a light, 1.10-m wire. If the ball is displaced \(8.50^{\circ}\) from the vertical and released, find the amplitude and period of the resulting swings of this pendulum.

A 2.00 -kg, frictionless block is attached to an ideal spring with force constant 300 \(\mathrm{N} / \mathrm{m} .\) At \(t=0\) the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 \(\mathrm{m} / \mathrm{s} .\) Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

Inside a NASA test vehicle, a \(3.50-\) kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The force constant of the spring is 225 \(\mathrm{N} / \mathrm{m}\) . The vehicle has a steady acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2},\) and the ball is not oscillating. Suddenly, when the vehicle's speed has reached \(45.0 \mathrm{m} / \mathrm{s},\) its engines turn off, thus eliminating its acceleration but not its velocity. Find (a) the amplitude and (b) the frequency of the resulting oscillations of the ball. (c) What will be the ball's maximum speed relative to the vehicle?

A 1.50-kg ball and a 2.00 -kg ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant \(165 \mathrm{N} / \mathrm{m},\) and the system is vibrating vertically with amplitude 15.0 \(\mathrm{cm} .\) The glue connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.

BIO Weighing a Virus. In February \(2004,\) scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 30 \(\mathrm{nm}\) long with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \(\left(f_{\mathrm{S}+\mathrm{v}}\right)\) to the frequency without the virus \(\left(f_{\mathrm{S}}\right)\) is given by the formula \(\frac{f_{\mathrm{S}+\mathrm{V}}}{f_{\mathrm{S}}}=\frac{1}{\sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)}}\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that is not necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of \(2.10 \times 10^{-16} \mathrm{g}\) and a frequency of \(2.00 \times 10^{15} \mathrm{Hz}\) without the virus and \(2.87 \times 10^{14} \mathrm{Hz}\) with the virus. What is the mass of the virus, in grams and in femtograms?
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