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As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. (a) What applies the force to the person to create the torque causing this acceleration? What is the direction of this force? (b) According to Newton's actionreaction law, what can you say about the direction of the force applied to the carousel by the person and about the nature (clockwise or counterclockwise) of the torque that it creates? (c) Does the torque identified in part (b) increase or decrease the angular speed of the carousel?

Short Answer

Expert verified
(a) The carousel exerts a radial and tangential force on the person. (b) The person exerts an opposite tangential force, creating clockwise torque. (c) This torque decreases the angular speed of the carousel.

Step by step solution

01

Identify Forces on the Person

When the person grabs onto the carousel bar, the centrifugal force due to the carousel's rotation acts on the person. This force is directed radially outward from the center of the carousel. However, the significant force applied here is the static frictional force from the bar on the person's hands, pulling the person in a tangential direction (aligned with the carousel's rotation) that causes them to accelerate and match the carousel's angular velocity.
02

Newton's Third Law

According to Newton's third law, every action has an equal and opposite reaction. As the static frictional force from the carousel acts on the person, the person applies an equal force in the opposite direction on the carousel. This force, when acting tangentially, creates a torque on the carousel.
03

Direction and Nature of the Force and Torque on the Carousel

The direction of the force applied by the person onto the carousel is opposite to the direction of movement of the frictional force initially exerted by the carousel on the person (i.e., it is tangential but opposite the rotation of the carousel). Hence, the torque created by this force on the carousel is clockwise, opposing the initial counterclockwise rotation.
04

Effect on the Angular Speed of the Carousel

Since the torque applied by the person on the carousel is clockwise, and the carousel's initial rotation is counterclockwise, this torque acts to decrease the overall angular speed of the carousel due to its opposite directional nature.

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Key Concepts

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

Newton's Third Law
Newton's Third Law states that for every action, there is an equal and opposite reaction. This fundamental principle of physics is particularly important when understanding forces and torques involved in rotational systems like the carousel problem. When the person grasps the bar to climb aboard, the static frictional force from the bar acts on them, causing a tangential acceleration.

Under Newton's Third Law, the carousel experiences a force of equal magnitude but in the opposite direction, applied by the person back onto the carousel. This reciprocal force results in a torque that affects the carousel's rotation.
  • The action: Frictional force pulling the person tangentially.
  • The reaction: Equal force applied by the person on the carousel.
Both forces are crucial, one creating acceleration for the person and the other affecting the carousel's rotational motion by producing a torque.
Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path, directed towards the center of rotation. When considering the motion of the person on the carousel, this force is paramount in maintaining circular motion.

As the carousel spins, any object on it, including a person climbing on, experiences a centripetal force keeping them on the circular path. For someone standing at the edge of the carousel, this force is furnished by the friction between their hands and the carousel's outer bars.
  • The direction of centripetal force is always inward, towards the center of the rotation.
  • Without sufficient centripetal force, the object would fly off tangentially due to inertia.
Understanding centripetal force helps visualize how the carousel and the person interact to maintain the path, preventing outward slipping.
Angular Acceleration
Angular acceleration refers to the rate of change of angular velocity with respect to time. In this scenario, when the person grabs onto the carousel, they start from rest and undergo angular acceleration to reach the carousel's speed. This is caused by the tangential force exerted by the carousel's bars.

This acceleration results in a change in the person's angular speed until it matches that of the carousel. The torque applied by the person's hands generates this acceleration.
  • Greater the applied torque, the greater the angular acceleration.
  • The initial angular speed is zero, increasing as the person matches the carousel's speed.
This concept illustrates how forces convert into motion on the carousel, demonstrating the transition from zero to nonzero angular speed for the person.

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Most popular questions from this chapter

A flat uniform circular disk (radius \(=2.00 \mathrm{~m}\), mass \(=1.00 \times 10^{2} \mathrm{~kg}\) ) is initially stationary. The disk is free to rotate in the horizontal plane about a frictionless axis perpendicular to the center of the disk. A \(40.0-\mathrm{kg}\) person, standing \(1.25 \mathrm{~m}\) from the axis, begins to run on the disk in a circular path and has a tangential speed of \(2.00 \mathrm{~m} / \mathrm{s}\) relative to the ground. Find the resulting angular speed of the disk (in \(\mathrm{rad} / \mathrm{s}\) ) and describe the direction of the rotation.

Concept Questions As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. (a) What applies the force to the person to create the torque causing this acceleration? What is the direction of this force? (b) According to Newton's actionreaction law, what can you say about the direction of the force applied to the carousel by the person and about the nature (clockwise or counterclockwise) of the torque that it creates? (c) Does the torque identified in part (b) increase or decrease the angular speed of the carousel? Problem The carousel has a radius of \(1.50 \mathrm{~m}\), an initial angular speed of \(3.14 \mathrm{rad} / \mathrm{s}\), and a moment of inertia of \(125 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The mass of the person is \(40.0 \mathrm{~kg}\). Find the final angular speed of the carousel after the person climbs aboard. Verify that your answer is consistent with your answers to the Concept Questions.

A cylindrically shaped space station is rotating about the axis of the cylinder to create artificial gravity. The radius of the cylinder is \(82.5 \mathrm{~m}\). The moment of inertia of the station without people is \(3.00 \times 10^{9} \mathrm{~kg} \cdot \mathrm{m}^{2}\). Suppose 500 people, with an average mass of \(70.0 \mathrm{~kg}\) each, live on this station. As they move radially from the outer surface of the cylinder toward the axis, the angular speed of the station changes. What is the maximum possible percentage change in the station's angular speed due to the radial movement of the people?

A pair of forces with equal magnitudes, opposite directions, and different lines of action is called a "couple." When a couple acts on a rigid object, the couple produces a torque that does not depend on the location of the axis. The drawing shows a couple acting on a tire wrench, each force being perpendicular to the wrench. Determine an expression for the torque produced by the couple when the axis is perpendicular to the tire and passes through (a) point \(\mathrm{A},\) (b) point \(\mathrm{B}\), and (c) point C. Express your answers in terms of the magnitude \(F\) of the force and the length \(L\) of the wrench.

A baggage carousel at an airport is rotating with an angular speed of \(0.20 \mathrm{rad} / \mathrm{s}\) when the baggage begins to be loaded onto it. The moment of inertia of the carousel is 1500 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Ten pieces of baggage with an average mass of \(15 \mathrm{~kg}\) each are dropped vertically onto the carousel and come to rest at a perpendicular distance of \(2.0 \mathrm{~m}\) from the axis of rotation. (a) Assuming that no net external torque acts on the system of carousel and baggage, find the final angular speed. (b) In reality, the angular speed of a baggage carousel does not change. Therefore, what can you say qualitatively about the net external torque acting on the system?

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