/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A thin uniform rod (mass \(=0.50... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A thin uniform rod (mass \(=0.50 \mathrm{~kg}\) ) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period of \(1.5\) s and an angular amplitude of \(10^{\circ} .\) (a) What is the length of the rod? (b) What is the maximum kinetic energy of the rod as it swings?

Short Answer

Expert verified
The length of the rod is 0.56 m, and the maximum kinetic energy is 0.014 J.

Step by step solution

01

Find the Moment of Inertia

The moment of inertia for a uniform rod about one end is given by the formula \( I = \frac{1}{3} m L^2 \), where \( m \) is the mass of the rod and \( L \) is its length. We will use this formula in conjunction with the known period to find the length.
02

Use the Pendulum Formula

For a physical pendulum, the period \( T \) is given by the formula \( T = 2 \pi \sqrt{\frac{I}{mgh}} \), where \( g = 9.8 \, \mathrm{m/s^2} \) is the acceleration due to gravity, and \( h = \frac{L}{2} \) is the distance from the pivot to the center of mass of the rod. Rearrange this formula to solve for \( L \).
03

Solve for Length \( L \)

Substitute the given period \( T = 1.5 \, s \), mass \( m = 0.50 \, kg \), and \( g = 9.8 \, \mathrm{m/s^2} \) into the formula and solve for \( L \): \(\begin{align*}T &= 2\pi \sqrt{\frac{\frac{1}{3}mL^2}{mg \frac{L}{2}}} \1.5 &= 2\pi \sqrt{\frac{2L}{3g}} \\left(\frac{1.5}{2\pi}\right)^2 &= \frac{2L}{3 \times 9.8} \L &= \left(\frac{1.5}{2\pi}\right)^2 \cdot \frac{3 \times 9.8}{2} \approx 0.56 \, \mathrm{m}.\end{align*}\)
04

Find the Maximum Angular Speed

At maximum speed, the kinetic energy is also maximum and occurs when the rod passes through the vertical position. Use the small-angle approximation for angular oscillations \( \omega_{\max} = \sqrt{ \frac{g}{L} } \cdot \theta_{\max} \), where \( \theta_{\max} = 10^{\circ} = \frac{\pi}{18} \approx 0.175 \, \text{rad} \).
05

Calculate Maximum Kinetic Energy

The kinetic energy is given by \( KE = \frac{1}{2} I \omega_{\max}^2 \). Use the moment of inertia from Step 1 and the angular speed from Step 4 to find \( KE \): \(\begin{align*}I &= \frac{1}{3} \times 0.50 \times (0.56)^2 \approx 0.0524 \, \mathrm{kg \cdot m^2} \\omega_{\max} &= \sqrt{ \frac{9.8}{0.56} } \times 0.175 \approx 0.73 \, \mathrm{rad/s} \KE &= \frac{1}{2} \times 0.0524 \times 0.73^2 \approx 0.014 \, \mathrm{J}.\end{align*}\)
06

Summary of Solutions

The solutions to the problems are as follows: (a) the length of the rod is approximately \( 0.56 \, \mathrm{m} \). (b) The maximum kinetic energy of the rod is approximately \( 0.014 \, \mathrm{J} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Moment of Inertia
The moment of inertia is a measure of how difficult it is to change the rotational motion of an object. Imagine a spinning rod or a wheel. The bigger or more spread out the mass, the harder it is to get it spinning or to stop it. For a thin uniform rod swinging about an axis at one of its ends, the moment of inertia is calculated using the formula:
  • \[ I = \frac{1}{3} m L^2 \]
Here, \( m \) stands for the mass of the rod, and \( L \) is the length of the rod. Every point of the rod contributes to its moment of inertia based on its distance from the pivot point. The further the mass is from the pivot, the higher the moment of inertia, which means it's tougher to accelerate or decelerate. This concept is crucial when calculating how objects will behave when they rotate around fixed points like in this exercise.
Kinetic Energy
Kinetic energy is the energy possessed by an object because of its motion. For a swinging pendulum, this energy is at its peak when the pendulum is at its lowest point during the swing. Understanding the kinetic energy helps us gauge how quickly an object moves when released from rest.For rotational motion, the formula for kinetic energy involves the moment of inertia and angular velocity:
  • \[ KE = \frac{1}{2} I \omega_{\max}^2 \]
In this scenario, \( I \) represents the moment of inertia we calculated earlier, and \( \omega_{\max} \) is the maximum angular speed the rod achieves. The maximum angular speed can be estimated using the following equations:
  • \[ \omega_{\max} = \sqrt{\frac{g}{L}} \times \theta_{\max} \]
This equation uses the gravitational constant \( g \) and the length of the pendulum \( L \) to ensure the object swings naturally back and forth under the influence of gravity. Finally, when calculating, remember to convert angular amplitude \( \theta_{\max} \) from degrees to radians, as this is the unit required for calculations involving angular speed. The knowledge of kinetic energy is useful in predicting how fast or slow a pendulum can swing at its maximum speed point.
Pendulum Period
The period of a pendulum is the time it takes to complete one full swing, back and forth. For a physical pendulum like a rod, understanding the period can help us design systems that keep time or respond to regular intervals.The period \( T \) for such a pendulum is given by:
  • \[ T = 2\pi \sqrt{\frac{I}{mgh}} \]
Here, \( I \) is the moment of inertia, \( m \) is the mass of the rod, \( g \) is the acceleration due to gravity, and \( h \) represents the distance from the pivot to the rod's center of mass. In a rod, this is halfway along its length or \( \frac{L}{2} \).By rearranging this formula, we can solve for the pendulum’s length \( L \) using known values for the gravitational force and the period, allowing us to calculate the unknowns related to the rod’s motion.Understanding the pendulum period is vital for scenarios where timing is critical, like pendulum clocks or even for certain musical instruments requiring timed movements to produce sound.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An oscillator consists of a block attached to a spring \((k=400 \mathrm{~N} / \mathrm{m})\). At some time \(t\), the position (measured from the system's equilibrium location), velocity, and acceleration of the block are \(x=0.100 \mathrm{~m}, v=-13.6 \mathrm{~m} / \mathrm{s}\), and \(a=-123 \mathrm{~m} / \mathrm{s}^{2} .\) Calcul- ate (a) the frequency of oscillation, (b) the mass of the block, and (c) the amplitude of the motion.

A \(5.00 \mathrm{~kg}\) object on a horizontal frictionless surface is attached to a spring with \(k=1000 \mathrm{~N} / \mathrm{m}\). The object is displaced from equilibrium \(50.0 \mathrm{~cm}\) horizontally and given an initial velocity of \(10.0 \mathrm{~m} / \mathrm{s}\) back toward the equilibrium position. What are (a) the motion's frequency, (b) the initial potential energy of the block-spring system, (c) the initial kinetic energy, and (d) the motion's amplitude?

The function \(x=(6.0 \mathrm{~m}) \cos [(3 \pi \mathrm{rad} / \mathrm{s}) t+\pi / 3 \mathrm{rad}]\) gives the simple harmonic motion of a body. At \(t=2.0 \mathrm{~s}\), what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?

A spider can tell when its web has captured,say, a fly because the fly's thrashing causes the web threads to oscillate. A spider can even determine the size of the fly by the frequency of the oscillations. Assume that a fly oscillates on the capture thread on which it is caught like a block on a spring. What is the ratio of oscillation frequency for a fly with mass \(m\) to a fly with mass \(2.5 \mathrm{~m}\) ?

A simple harmonic oscillator consists of a block of mass \(2.00\) kg attached to a spring of spring constant \(100 \mathrm{~N} / \mathrm{m}\). When \(t=1.00\) s, the position and velocity of the block are \(x=0.129 \mathrm{~m}\) and \(v=\) \(3.415 \mathrm{~m} / \mathrm{s}\). (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at \(t=0 \mathrm{~s}\) ?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.