/*! 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 45 A light rod of length \(\ell=1.0... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A light rod of length \(\ell=1.00 \mathrm{~m}\) rotates about an axis perpendicular to its length and passing through its center as in Figure P8.45. Two particles of masses \(m_{1}=4.00 \mathrm{~kg}\) and \(m_{2}=3.00 \mathrm{~kg}\) are connected to the ends of the rod. (a) Neglecting the mass of the rod, what is the system's kinetic energy when its angular speed is \(2.50 \mathrm{rad} / \mathrm{s}\) ? (b) Repeat the problem, assuming the mass of the rod is taken to be \(2.00 \mathrm{~kg}\).

Short Answer

Expert verified
The kinetic energy of the system is 2.734375 J when the mass of the rod is neglected, and it's 4.296875 J when the mass of the rod is considered.

Step by step solution

01

Calculating the moment of inertia in Part (a)

Firstly, the moment of inertia when the mass of the rod is neglected is calculated as the sum of the moments of inertia of each particle. For each particle, the moment of inertia is \( I = m * r^2 \), where \( r = \ell / 2 \) is the distance from the rotation axis. Hence, the total moment of inertia is \( I_{total} = m_1 * (\ell / 2)^2 + m_2 * (\ell / 2)^2 = 4 * (1 / 2)^2 + 3 * (1 / 2)^2 = 0.5 + 0.375 = 0.875 \, kg \cdot m^2 \)
02

Calculating the kinetic energy in Part (a)

Using the calculated moment of inertia and the given angular speed \( \omega = 2.5 \, rad/s \), we calculate the kinetic energy as \( KE = 0.5 * I_{total} * \omega^2 = 0.5 * 0.875 * (2.5)^2 = 2.734375 \, J \)
03

Calculating the moment of inertia in Part (b)

The moment of inertia when the mass of the rod is considered includes the contribution from the rod as \( I_{rod} = m_{rod} * (\ell / 2)^2 \). So the total moment of inertia is \( I_{total} = m_1 * (\ell / 2)^2 + m_2 * (\ell / 2)^2 + I_{rod} = 0.875 + 2.0 * (1 / 2)^2 = 0.875 + 0.5 = 1.375 \, kg \cdot m^2 \)
04

Calculating the kinetic energy in Part (b)

Using the calculated moment of inertia for part (b) and the given angular speed, the kinetic energy is \( KE = 0.5 * I_{total} * \omega^2 = 0.5 * 1.375 * (2.5)^2 = 4.296875 \, J \)

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
Understanding the moment of inertia is essential in dealing with rotational motion in physics. It's the rotational equivalent of mass for linear motion. The moment of inertia depends on the distribution of mass around an axis and determines how much torque is needed for a given angular acceleration. In the problem above, the moment of inertia, represented by the symbol 'I', is calculated using the formula for point masses, which is I = m * r^2.

Let's break down this equation: 'm' stands for the mass of the object, and 'r' is the distance from the rotation axis to the point mass. In the case of multiple point masses, you add up their individual moments of inertia to find the total moment of inertia. This gives you a complete picture of how the system's mass impact its rotational characteristics, even if some parts of the system, like the rod in our exercise, are considered to have negligible mass.

If the mass of a rod or another object should be considered, however, there are specific formulas based on the object's shape and mass distribution. An example of this is a uniform rod rotating about its center, which has a different moment of inertia than if it were rotating about one end. Accurately calculating the moment of inertia is crucial, as it directly affects the rotational kinetic energy of a system.
Angular Speed
Angular speed is a measure of how fast an object rotates or revolves relative to another point, commonly the center of rotation. It's denoted by the Greek letter 'ω' (omega) and is defined by how many radians an object covers per unit of time. Radians are a way to measure angles based on the radius of a circle, with one full revolution equating to 2π radians.

In our exercise, the angular speed is given as 2.50 rad/s, meaning the system completes 2.50 radians of its circular path every second. Understanding angular speed is vital for solving problems involving rotational kinetic energy, as it plays a significant part in determining how much kinetic energy an object in rotation has. This is due to the fact that kinetic energy is directly proportional to the square of the angular speed, highlighted in the equation KE = 0.5 * I * ω^2. This quadratic relationship shows that even a small increase in angular speed results in a relatively large increase in the rotational kinetic energy.
Physics Calculations
Physics calculations often involve synthesizing formulas, understanding concepts, and applying mathematical operations to solve a problem. For instance, when calculating the kinetic energy of rotating objects, one needs to understand both the physical concepts and be adept with algebra.

In the given solution steps, we first calculate the moment of inertia and then utilize it to compute the kinetic energy. Such calculations require precision and attention to units to ensure accuracy. It's important to be systematic: first, identify known quantities, second, select the appropriate formulas, then substitute the values carefully, and finally, compute the result.

When an additional mass is considered, such as the mass of the rod in part (b), it's crucial to correctly include that in the moment of inertia calculation. The systematic approach and careful execution can prevent mistakes in complicated problems and is an invaluable skill in physics and many other sciences.

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

QC A \(40.0-\mathrm{kg}\) child stands at one end of a \(70.0-\mathrm{kg}\) boat that is \(4.00 \mathrm{~m}\) long (Fig. P8.69). The boat is initially \(3.00 \mathrm{~m}\) from the pier. The child notices a turtle on a rock beyond the far end of the boat and proceeds to walk to that end to catch the turtle. (a) Neglecting friction between the boat and water, describe the motion of the system (child plus boat). (b) Where will the child be relative to the pier when he reaches the far end of the boat? (c) Will he catch the turtle? (Assume that he can reach out \(1.00 \mathrm{~m}\) from the end of the boat.)

S This is a symbolic version of problem 72. Two astronauts (Fig. P8.72), each having a mass \(M\), are connected by a rope of length \(d\) having negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed \(v\). (a) Calculate the magnitude of the angular momentum of the system by treating the astronauts as particles. (b) Calculate the rotational energy of the system. By pulling on the rope, the astronauts shorten the distance between them to \(d / 2\). (c) What is the new angular momentum of the system? (d) What are their new speeds? (e) What is the new rotational energy of the system? (f) How much work is done by the astronauts in shortening the rope?

An airliner lands with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\). Each wheel of the plane has a radius of \(1.25 \mathrm{~m}\) and a moment of inertia of \(110 \mathrm{~kg} \cdot \mathrm{m}^{2}\). At touchdown, the wheels begin to spin under the action of friction. Each wheel supports a weight of \(1.40 \times 10^{4} \mathrm{~N}\), and the wheels attain their angular speed in \(0.480 \mathrm{~s}\) while rolling without slipping. What is the coefficient of kinetic friction between the wheels and the runway? Assume that the speed of the plane is constant.

QIC A \(0.00500-\mathrm{kg}\) bullet traveling horizontally with a speed of \(1.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\) enters an \(18.0-\mathrm{kg}\) door, embedding itself \(10.0 \mathrm{~cm}\) from the side opposite the hinges as in Figure P8.56. The \(1.00-\mathrm{m}\)-wide door is free to swing on its hinges. (a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? Explain. (b) Is mechanical energy con-served in this collision? Answer without doing a calculation. (c) At what angular speed does the door swing open immediately after the collision? (The door has the same moment of inertia as a rod with axis at one end.) (d) Calculate the energy of the door-bullet system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.

M A \(150-\mathrm{kg}\) merry-go-round in the shape of a uniform, solid, horizontal disk of radius \(1.50 \mathrm{~m}\) is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of \(0.500 \mathrm{rev} / \mathrm{s}\) in \(2.00 \mathrm{~s}\) ?

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.