/*! 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 60 A thin rod of length \(0.75 \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin rod of length \(0.75 \mathrm{~m}\) and mass \(0.42 \mathrm{~kg}\) is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed \(4.0 \mathrm{rad} / \mathrm{s}\). Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

Short Answer

Expert verified
(a) Kinetic energy is 0.63 J. (b) Center of mass rises 0.153 m above its lowest position.

Step by step solution

01

Understand the Problem

We have a rod acting as a pendulum. We need to calculate two things at the lowest position: the kinetic energy of the rod and how high the center of mass rises above that lowest point.
02

Find Kinetic Energy at Lowest Position

The formula for rotational kinetic energy is \( KE = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia of the rod and \( \omega \) is the angular speed. For a thin rod pivoted about one end, \( I = \frac{1}{3} m L^2 \). Substitute the values: \( m = 0.42 \text{ kg} \), \( L = 0.75 \text{ m} \), and \( \omega = 4.0 \text{ rad/s} \). Calculate \( I \) and then \( KE \).
03

Calculate Moment of Inertia

Use the formula \( I = \frac{1}{3} m L^2 \). Substitute the given values: \( m = 0.42 \text{ kg} \) and \( L = 0.75 \text{ m} \). Compute \( I = \frac{1}{3} \times 0.42 \times (0.75)^2 = 0.07875 \text{ kg} \cdot \text{m}^2 \).
04

Calculate Kinetic Energy

Now substitute \( I = 0.07875 \text{ kg} \cdot \text{m}^2 \) and \( \omega = 4.0 \text{ rad/s} \) into the kinetic energy formula: \( KE = \frac{1}{2} \times 0.07875 \times (4.0)^2 = 0.63 \text{ J} \).
05

Use Energy Conservation to Find Height

The initial potential energy at the highest point is converted to kinetic energy at the lowest point. Use the formula \( KE = mgh \) to find the height \( h \): \( 0.63 = 0.42 \times 9.8 \times h \). Solve for \( h \).
06

Solve for Height

Rearrange the formula \( h = \frac{KE}{mg} \): \( h = \frac{0.63}{0.42 \times 9.8} = 0.153 \text{ m} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In the context of rotational motion, like the swinging of a pendulum, kinetic energy depends on the object's moment of inertia and its angular speed. The formula to calculate rotational kinetic energy is:
  • \( KE = \frac{1}{2} I \omega^2 \)
where \( KE \) is the kinetic energy, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. This indicates that kinetic energy increases as the speed of the object increases, or as the distribution of an object's mass increases, as represented by moment of inertia.
The kinetic energy of the rod at its lowest point reflects the conversion of all potential energy to motion, signifying the importance of both factors in rotational systems.
Moment of Inertia
Moment of inertia is a measure of an object's resistance to changes in its rotation. It depends not only on the mass of the object but also on how that mass is distributed relative to the axis of rotation. For a thin rod rotating about one end, the specific formula used is:
  • \( I = \frac{1}{3} m L^2 \)
In this equation, \( m \) represents the mass of the rod, and \( L \) is the length of the rod. This tells us that the further the mass is from the pivot point, the larger the moment of inertia, making it harder for the object to start or stop rotating.
Understanding moment of inertia is crucial for analyzing rotational motion because it affects how the object accelerates as well as its kinetic energy.
Pendulum Motion
A pendulum exemplifies simple harmonic motion where potential energy and kinetic energy continuously convert into one another. As a pendulum swings downward from its highest point, potential energy decreases while kinetic energy increases, reaching its maximum at the lowest point.
This back-and-forth motion is a balance between gravitational force pulling the pendulum towards its lowest point and the inertia that wants to keep it moving past that point.
Pendulum motion in objects like our rod can be studied using the principles of conservation of energy to predict behavior over time.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed—only transformed from one form to another. In this exercise, the swing of the rod illustrates this principle beautifully. At the highest point of the swing, all energy is potential due to height. At the lowest point, all this potential energy converts to kinetic energy.To find how high the center of mass of the rod rises, we make use of this principle in the equation:
  • \( KE = mgh \)
where \( KE \) is the kinetic energy at the lowest point, \( m \) is the mass, \( g \) is the gravitational acceleration, and \( h \) is the height above the lowest point. Knowing that the energy at the lowest point is 0.63 Joules, and solving, the rod's center of mass rises approximately 0.153 meters.
This conservation principle is key to understanding and solving many physics problems involving motion.

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Most popular questions from this chapter

A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bumps in the groove run into the stylus, causing it to oscillate. The equipment converts those oscillations to electrical signals and then to sound. Suppose that a record turns at the rate of \(33 \frac{1}{3}\) rev/min, the groove being played is at a radius of \(10.0 \mathrm{~cm}\), and the bumps in the groove are uniformly separated by \(1.75 \mathrm{~mm}\). At what rate (hits per second) do the bumps hit the stylus?

A gyroscope flywheel of radius \(2.83 \mathrm{~cm}\) is accelerated from rest at \(14.2 \mathrm{rad} / \mathrm{s}^{2}\) until its angular speed is \(2760 \mathrm{rev} / \mathrm{min} .\) (a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process? (b) What is the radial acceleration of this point when the flywheel is spinning at full speed? (c) Through what distance does a point on the rim move during the spin-up?

A flywheel with a diameter of \(1.20 \mathrm{~m}\) is rotating at an angular speed of 200 rev/min. (a) What is the angular speed of the flywheel in radians per second? (b) What is the linear speed of a point on the rim of the flywheel? (c) What constant angular acceleration (in revolutions per minute- squared) will increase the wheel's angular speed to 1000 rev/min in \(60.0 \mathrm{~s}\) ? (d) How many revolutions does the wheel make during that \(60.0 \mathrm{~s}\) ?

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The angular position of a point on a rotating wheel is given by \(\theta=2.0+4.0 t^{2}+2.0 t^{3}\), where \(\theta\) is in radians and \(t\) is in seconds. At \(t=0\), what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at \(t=4.0 \mathrm{~s}\) ? (d) Calculate its angular acceleration at \(t=2.0 \mathrm{~s}\). (e) Is its angular acceleration constant?

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