/*! 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 66 A merry-go-round rotates at the ... [FREE SOLUTION] | 91Ó°ÊÓ

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A merry-go-round rotates at the rate of \(0.20 \mathrm{rev} / \mathrm{s}\) with an \(80-\mathrm{kg}\) man standing at a point \(2.0 \mathrm{~m}\) from the axis of rotation. (a) What is the new angular speed when the man walks to a point \(1.0 \mathrm{~m}\) from the center? Assume that the merry-go-round is a solid \(25-\mathrm{kg}\) cylinder of radius \(2.0 \mathrm{~m} .\) (b) Calculate the change in kinetic energy due to the man's movement. How do you account for this change in kinetic energy?

Short Answer

Expert verified
The new angular speed when the man walks to a point 1.0 m from the center is approximately \(0.233 \mathrm{rev} / \mathrm{s}\). The change in kinetic energy due to the man's movement is approximately \( - 137 \mathrm{J}\). The negative sign indicates that energy is lost from the system, accounted for by the work done by the man in moving towards the center.

Step by step solution

01

Determine Initial Moment of Inertia and Calculate Angular Momentum

First, calculate the initial moment of inertia of the system. The merry-go-round can be considered as a solid cylinder whose moment of inertia \(I_1\) is given by \(0.5MR^2\) where \(M\) is the mass and \(R\) is the radius. For the man, it can be treated as a point mass so the moment of inertia \(I_m1\) is \(mr^2\) where \(m\) is the mass and \(r\) is the distance from the axis of rotation. So the total moment of inertia \(I_1\) will be \(0.5MR^2 + mr^2\). The initial angular momentum \(L_1\) will be the product of this moment of inertia and angular speed- \(L_1 = I_1ω_1\).
02

Calculate Final Moment of Inertia and Equate Angular Momenta

Next, calculate the final moment of inertia when the man moves. The man's new moment of inertia \(I_m2\) will now be \(mr^2\) with the new \(r\). The merry-go-round's moment of inertia won't change, so the total final moment of inertia \(I_2\) is \(0.5MR^2 + mr^2\). As per conservation of angular momentum, \(L_1 = L_2\). So, \(I_1ω_1 = I_2ω_2\), from which you can solve for the new angular speed \(ω_2\).
03

Calculate Initial and Final Kinetic Energy and Find The Change

Calculate initial kinetic energy of the system \(KE_1\) using the formula \(\frac{1}{2}I_1ω_1^2\). Similarly, calculate final kinetic energy \(KE_2\) with the formula \(\frac{1}{2}I_2ω_2^2\). The change in kinetic energy will then be \(∆KE = KE_2 - KE_1\). This change in kinetic energy is due to the work done by the man in walking towards the center.

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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 is a property of a physical object that quantifies how difficult it is to change its angular velocity. Think of it like the rotational counterpart to mass in linear motion. The larger the moment of inertia, the more torque you will need to speed up or slow down its rotation. For different shapes, there are different formulas to calculate this value. For example, a solid cylinder (like our merry-go-round) has a moment of inertia given by the equation \( I = 0.5MR^2 \) where \( M \) is the mass of the cylinder, and \( R \) is the radius. When a mass such as the man moves closer to the axis, his contribution to the overall moment of inertia changes because his distance to the axis changes, according to the equation \( I_m = mr^2 \) where \( m \) is the mass of the man and \( r \) is the distance from the rotational axis.
angular speed
Angular speed describes how quickly an object rotates or revs around an internal axis. It's measured in radians per second, but in everyday situations, like a merry-go-round, you might see it as revolutions per second. Once you know the angular speed, you can determine how fast someone on the edge of the merry-go-round is moving, even though they are simply standing in place. As our textbook example illustrates, if the man moves closer to the center, the system's angular momentum must be conserved. This means the merry-go-round needs to adjust its spinning speed to compensate for the man's movement as to keep the overall angular momentum constant.
kinetic energy change
Kinetic energy in the context of rotational motion is the energy an object possesses due to its rotation, which is given by the equation \( KE = \frac{1}{2}I\omega^2 \) where \( I \) is the moment of inertia and \( \omega \) is the angular speed. When conditions of the rotational system change, like our man moving towards the center, the kinetic energy also changes. This shift reflects the work done or energy transferred within the system. In our merry-go-round scenario, the man walking inwards is doing work against the centrifugal force, causing the system's kinetic energy to adjust.
rotational dynamics
Rotational dynamics is the branch of physics that describes the motion of objects rotating around an axis. It involves the concepts of torque, moment of inertia, and angular momentum. The principles governing these rotational quantities are analogous to the forces, mass, and momentum found in linear dynamics. Conservation of angular momentum, a key rule in this field, dictates that if no external torques act on a system, the total angular momentum will remain constant. This is central to the textbook problem, where the man moving towards the center of the merry-go-round does not alter the system's total angular momentum but affects its angular speed and kinetic energy, showcasing the intricate balance in a closed rotational system.

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Most popular questions from this chapter

A playground merry-go-round of radius \(2.00 \mathrm{~m}\) has a moment of inertia \(I=275 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and is rotating about a frictionless vertical axle. As a child of mass \(25.0 \mathrm{~kg}\) stands at a distance of \(1.00 \mathrm{~m}\) from the axle, the system (merrygo-round and child) rotates at the rate of \(14.0 \mathrm{rev} / \mathrm{min}\). The child then proceeds to walk toward the edge of the merry-go-round. What is the angular speed of the system when the child reaches the edge?

A potter's wheel having a radius of \(0.50 \mathrm{~m}\) and a moment of inertia of \(12 \mathrm{~kg} \cdot \mathrm{m}^{2}\) is rotating freely at \(50 \mathrm{rev} / \mathrm{min}\). The potter can stop the wheel in \(6.0 \mathrm{~s}\) by pressing a wet rag against the rim and exerting a radially inward force of \(70 \mathrm{~N}\). Find the effective coefficient of kinetic friction between the wheel and the wet rag.

A beam resting on two pivots has a length of \(L=\) \(6.00 \mathrm{~m}\) and mass \(M=90.0 \mathrm{~kg}\). The pivot under the left end exerts a normal force \(n_{1}\) on the beam, and the second pivot placed a distance \(\ell=4.00 \mathrm{~m}\) from the left end exerts a normal force \(n_{2} .\) A woman of mass \(m=55.0 \mathrm{~kg}\) steps onto the left end of the beam and begins walking to the right as in Figure \(\mathrm{P} 8.12\). The goal is to find the woman's position when the beam begins to tip. (a) Sketch a free-body diagram, labeling the gravitational and normal forces acting on the beam and placing the woman \(x\) meters to the right of the first pivot, which is the origin. (b) Where is the woman when the normal force \(n_{1}\) is the greatest? (c) What is \(n_{1}\) when the beam is about to tip? (d) Use the force equation of equilibrium to find the value of \(n_{2}\) when the beam is about to tip. (e) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip. (f) Check the answer to part (e) by computing torques around the first pivot point. Except for possible slight differences due to rounding, is the answer the same?

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 each cord and (b) the acceleration of each object after the objects have descended a distance \(h .\)

A simple pendulum consists of a small object of mass 3.0 kg hanging at the end of a \(2.0\) -m-long light string that is connected to a pivot point. (a) Calculate 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.

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