/*! 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 43 A horizontal \(800-\mathrm{N}\) ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 43

A horizontal \(800-\mathrm{N}\) merry-go-round of radius \(1.50 \mathrm{~m}\) is started from rest by a constant horizontal force of \(50.0 \mathrm{~N}\) applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go- round after \(3.00 \mathrm{~s}\). (Assume it is a solid cylinder.)

Short Answer

Expert verified
The kinetic energy of the merry-go-round after 3 seconds is calculated using the formulas from the steps above.

Step by step solution

01

Calculate Torque and Angular Acceleration

First, we calculate the torque (\(\tau\)) using the formula \(\tau = Fr\), where \(F\) is the force applied and \(r\) is the radius of the merry-go-round. We then use that torque to calculate the angular acceleration (\(\alpha\)) via the formula \(\tau = I\alpha\), where \(I\) is the moment of inertia of the merry-go-round.
02

Moment of Inertia

Calculate the moment of inertia for a solid cylinder, which is given by \(I = 0.5 MR^2\), where \(M\) is the mass of the merry-go-round and \(R\) is the radius. We can find \(M\) by rearranging the weight equation \(W = Mg\), since the weight of the merry-go-round is provided.
03

Calculate Final Angular Speed

The final angular speed (\(\omega\)) can be found by using the formula \(\omega = \omega_0 + \alpha t\), where \(\omega_0\) is the initial angular speed (which is zero as the merry-go-round starts from rest), \(\alpha\) is the angular acceleration, and \(t\) is the time, which is given as 3.00 s.
04

Calculate Kinetic Energy

Finally, we calculate the kinetic energy using the rotational kinetic energy formula \(K = 0.5 I \omega^2\), where \(I\) is the moment of inertia and \(\omega\) is the final angular speed.

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Key Concepts

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

Torque
Torque, often denoted as \(\tau\), is a measure of how much a force acting on an object causes that object to rotate. For any object, the torque is the product of the force applied \(F\) and the distance from the pivot point to the point where the force is applied \(r\), also known as the lever arm. The formula for torque is \(\tau = Fr\).

Applying this concept to the merry-go-round problem, we can understand how the horizontal force causes rotational movement. In the given situation, a tangential force is applied to the edge of the object, which is the point farthest from the axis of rotation, maximizing the effect of the torque. This tangential force is the starting point to understanding how the merry-go-round accelerates from rest and gains kinetic energy.
Angular Acceleration
Angular acceleration, symbolized as \(\alpha\), describes the rate of change of angular speed. It is the rotational analogue to linear acceleration and tells us how quickly an object is speeding up or slowing down in its rotational movement. When we have calculated the torque, we can find angular acceleration using the second part of the torque equation \(\tau = I\alpha\), where \(I\) is the moment of inertia.

In the example of the merry-go-round, knowing the torque and the moment of inertia, we can determine the angular acceleration. This reveals how rapidly the rotational speed of the merry-go-round increases over time due to the applied force. Angular acceleration is essential for understanding how the motion evolves from the initial application of force.
Moment of Inertia
The moment of inertia \(I\) is a property of a rotating object that quantifies its resistance to angular acceleration. Different shapes and mass distributions around the rotation axis yield different moments of inertia. For a solid cylinder, such as a merry-go-round, the moment of inertia is expressed by the formula \(I = 0.5 MR^2\), where \(M\) is the mass and \(R\) is the radius.

By understanding the moment of inertia, we comprehend why some objects are easier or harder to spin. In our exercise, calculating the moment of inertia of the merry-go-round is crucial as it directly impacts both the torque necessary for a given angular acceleration and the final kinetic energy of the rotating object.
Rotational Kinetic Energy
Rotational kinetic energy represents the energy of an object due to its rotation and is particularly relevant in systems involving spinning parts. For rotating objects, this energy is given by the formula \(K = 0.5 I \omega^2\), where \(I\) is the moment of inertia and \(\omega\) is the angular speed of the object. Just like with linear kinetic energy, the greater the rotation speed, the more kinetic energy an object has.

In reference to our merry-go-round, after finding the moment of inertia and the final angular speed, we are able to calculate how much kinetic energy it has gained after being subject to the force for three seconds. This calculation allows us to quantify the energy associated with its rotational movement, resultant from the applied force and inherent material properties.

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Most popular questions from this chapter

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 \(2.00 \mathrm{~s}\) ?

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?

A \(240-\mathrm{N}\) sphere \(0.20 \mathrm{~m}\) in radius rolls without slipping \(6.0 \mathrm{~m}\) down a ramp that is inclined at \(37^{\circ}\) with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?

A space station shaped like a giant wheel has a radius of \(100 \mathrm{~m}\) and a moment of inertia of \(5.00 \times 10^{8} \mathrm{~kg} \cdot \mathrm{m}^{2}\) A crew of 150 lives on the rim, and the station is rotating so that the crew experiences an apparent acceleration of \(1 g\) (Fig. P8.64).When 100 people move to the center of the station for a union meeting, the angular speed changes. What apparent acceleration is experienced by the managers remaining at the rim? Assume the average mass of a crew member is \(65.0 \mathrm{~kg}\).

A large grinding wheel in the shape of a solid cylinder of radius \(0.330 \mathrm{~m}\) is free to rotate on a frictionless, vertical axle. A constant tangential force of \(250 \mathrm{~N}\) applied to its edge causes the wheel to have an angular acceleration of \(0.940 \mathrm{rad} / \mathrm{s}^{2}\). (a) What is the moment of inertia of the wheel? (b) What is the mass of the wheel? (c) If the wheel starts from rest, what is its angular velocity after \(5.00 \mathrm{~s}\) have elapsed, assuming the force is acting during that time?
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