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Chapter 8: Problem 39

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

Short Answer

Expert verified
The constant force that must be exerted on the rope to bring the merry-go-round from rest to an angular speed of \(0.500\) rev/s in \(2.00\) s is \(176.84\) N.

Step by step solution

01

Finding Angular Acceleration

Firstly, we need to find the angular acceleration (伪) of the merry-go-round. The angular speed (蠅) can be converted from revolutions per second to radians per second using the conversion factor \(2蟺\) radians/revolution, to give \( 蠅 = 0.500 rev/s 脳 2蟺 rad/rev = 蟺 rad/s\). Angular acceleration (伪) can then be found using the formula \( 伪 = 蠅 / t \), where t = 2.0 s. Substituting the known values, we get \( 伪 = 蟺 rad/s / 2.0 s = 蟺/2 rad/s虏 \)
02

Calculating Moment of Inertia

Since the merry-go-round is in the shape of a uniform, solid, horizontal disk, its moment of inertia (I) can be found using the formula \( I = 0.5MR虏 \), where M = 150 kg is the mass of the merry-go-round, and R = 1.50 m is its radius. Substituting the known values, we get \( I = 0.5 脳 150 kg 脳 (1.50 m)虏 = 168.75 kg.m虏 \)
03

Finding Torque

Torque (蟿) can now be found by using the formula 蟿 = I伪. Substituting the known values, we get \( 蟿= 168.75 kg.m虏 脳 蟺/2 rad/s虏 = 265.26 N.m \)
04

Calculating Force Required

Finally, we can find the force (F) needed to set the merry-go-round in motion using the formula \( F = 蟿/R \), where R = 1.50 m is the radius of the merry-go-round. Substituting the known values, we get \( F = 265.26 N.m /1.50 m = 176.84 N \)

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Key Concepts

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

Moment of Inertia
The moment of inertia, often expressed as 'I', is a crucial aspect of rotational motion that represents the resistance an object has to changes in its rotational state. It depends not only on the mass of the object but also on how this mass is distributed with respect to the axis of rotation.

In our example of the merry-go-round, the moment of inertia is determined using the formula for a uniform solid disc, which is \( I = \frac{1}{2}MR^2 \). Here, 'M' signifies the mass, and 'R' is the radius of the disc. It's similar to mass in a linear motion, except it's all about how the mass is 'spread out' in rotation. The larger the moment of inertia, the more force you'll need to spin the object at the same rate as one with a smaller moment of inertia.
Rotational Motion
Rotational motion relates to the movement of an object around a center or an axis. All points in the body move in circles about the axis, and angular speed (\( \omega \)) highlights how fast these points are spinning. However, it's angular acceleration (\( \alpha \)) that quantifies how quickly the angular speed changes over time.

For the merry-go-round, angular acceleration was found using \( \alpha = \frac{\omega}{t} \), with 't' as the time to reach the angular speed \( \omega \). The acceleration shows how quickly the merry-go-round goes from stationary to spinning - like stepping on a car's gas pedal, except you're not going straight, you're spinning.
Torque
Torque can be thought of as the rotational equivalent of force. It's what causes an object to start rotating, stop rotating, or rotate more slowly. Technically, torque (\( \tau \)) is the cross product of the radius vector and the force vector, leading to the formula \( \tau = Fr \) where 'F' is the linear force applied perpendicularly at a distance, 'r', from the rotation axis.

In the case of our merry-go-round, we simplify this to \( \tau = I\alpha \), because we already know its moment of inertia (I) and we've calculated its angular acceleration (\( \alpha \)). The torque here represents the rotational influence of the pulling force applied viathe rope. Finally, knowing the torque and the radius of our disc, we're able to deduce the actual force required to pull on the rope to achieve the desired rotational effect.

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Most popular questions from this chapter

A student sits on a rotating stool holding two \(3.0-\mathrm{kg}\) objects. When his arms are extended horizontally, the objects are \(1.0 \mathrm{~m}\) from the axis of rotation and he rotates with an angular speed of \(0.75 \mathrm{rad} / \mathrm{s}\). The moment of inertia of the student plus stool is \(3.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and is assumed to be constant. The student then pulls in the objects horizontally to \(0.30 \mathrm{~m}\) from the rotation axis. (a) Find the new angular speed of the student. (b) Find the kinetic energy of the student before and after the objects are pulled in.

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 \(2.00 \mathrm{~kg}\).

S A uniform solid cylinder of mass \(M\) and radius \(R\) rotates on a frictionless horizontal axle (Fig. P8.81). Two objects with equal masses \(m\) hang from light cords wrapped around the cylinder. If the system is released from rest, find (a) the tension in eachcord and (b) the acceleration of each object after the objects have descended a distance \(h\).

QC A simple pendulum con- Figure p \(8.4\) sists of a small object of mass a \(2.0-\mathrm{m}\)-long light string that is connected to a pivot point. (a) Galculate the magnitude of the torque (due to the force of gravity) about this pivot point when the string makes a \(5.0^{\circ}\) angle with the vertical. (b) Does the torque increase or decrease as the angle increases? Explain.

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?
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