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Chapter 9: Problem 56

Example 14 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of \(125 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) When one person is standing at a distance of \(1.50 \mathrm{~m}\) from the center, the carousel has an angular velocity of \(0.600 \mathrm{rad} / \mathrm{s}\). However, as this person moves inward to a point located \(0.750 \mathrm{~m}\) from the center, the angular velocity increases to \(0.800 \mathrm{rad} / \mathrm{s}\). What is the person's mass?

Short Answer

Expert verified
The person's mass is approximately 27.78 kg.

Step by step solution

01

Understand the Concept

This problem involves the conservation of angular momentum. Since the carousel is frictionless, the angular momentum before the person moves inward must equal the angular momentum after the person moves inward.
02

Write the Angular Momentum Conservation Equation

The initial angular momentum is given by: \(L_i = (I + m r_i^2) \omega_i\), where \(I\) is the moment of inertia of the carousel, \(m\) is the mass of the person, \(r_i\) is the initial distance from the axis, and \(\omega_i\) is the initial angular velocity. The final angular momentum is \(L_f = (I + m r_f^2) \omega_f\), where \(r_f\) and \(\omega_f\) are the final distance and angular velocity respectively. According to the conservation law: \(L_i = L_f\).
03

Substitute and Simplify

Using the conservation of angular momentum: \[(I + m r_i^2) \omega_i = (I + m r_f^2) \omega_f\]. Substitute the known values: \[ (125 + m (1.5)^2)(0.600) = (125 + m (0.750)^2)(0.800) \].
04

Solve for the Person's Mass

First, simplify the equation: \[ (125 + 2.25m)(0.600) = (125 + 0.5625m)(0.800) \].Next, expand both sides: \[ 75 + 1.35m = 100 + 0.45m \].Rearrange to isolate \(m\) on one side: \[ 1.35m - 0.45m = 100 - 75 \]. Simplify this to \[ 0.90m = 25 \].Finally, solve for \(m\): \[ m = \frac{25}{0.9} \approx 27.78 \mathrm{~kg} \].
05

Verify Your Solution

Substitute \(m\) back into the original equation to ensure both sides are equal. This checks that no arithmetic error was made during calculations.

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Key Concepts

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

Understanding Moment of Inertia
The moment of inertia is a fundamental concept in physics, particularly when dealing with rotational motion. Think of it as the rotational equivalent of mass in linear motion. It quantifies how much "effort" is needed to change an object's rotational speed. For a rigid body, like the playground carousel in the exercise, the moment of inertia depends on its mass distribution relative to the axis of rotation.
In simpler terms:
  • Objects with mass further from the axis have a greater moment of inertia.
  • The moment of inertia of a system can change when its mass distribution changes (e.g., when the person moves closer to the center).
  • In the carousel problem, the moment of inertia for the system initially includes the carousel and the person standing 1.5 meters from the center.
When the person moves closer to the center, the overall distribution of mass changes, and so does the moment of inertia, impacting the rotational speed.
Exploring Angular Velocity
Angular velocity is an essential concept when discussing rotation, describing the rate of rotation around a central point. It's usually measured in radians per second. In a rotational system like the carousel, angular velocity tells us how quickly the carousel spins.
Key points about angular velocity:
  • It changes if an external torque is applied or if the moment of inertia changes (as in the carousel problem).
  • Angular velocity increases if the moment of inertia decreases, provided angular momentum is conserved.
  • In the given problem, as the person moves inward, the moment of inertia decreases, and thus, the angular velocity increases from 0.600 rad/s to 0.800 rad/s.
This increase in angular velocity when the person moves closer to the center showcases the direct relationship between moment of inertia and angular velocity under the law of conservation of angular momentum.
Solving Physics Problems with Conservation of Angular Momentum
Physics problems like the one given require understanding the principle of conservation of angular momentum. This principle states that if no external torque acts on a system, the total angular momentum remains constant. In the carousel scenario:
  • The system is isolated with no external forces acting, meaning the initial and final angular momentum must be equal.
  • The initial angular momentum includes contributions from both the carousel and the person at 1.5 meters.
  • After the person moves to 0.750 meters, the system adjusts to maintain the same angular momentum, resulting in a higher angular velocity.
Solving these types of problems:
  1. Identify what's included in the system's moment of inertia initially and finally.
  2. Write out the angular momentum conservation equation: initial equals final.
  3. Mathematically manipulate the equation to find unknown variables, like the person's mass in this problem.
Understanding these steps for solving conservation problems reveals how different factors, like position and velocity, interrelate to maintain the balance of angular forces.

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Most popular questions from this chapter

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{~kg})\) is rotating at an angular velocity of \(13.1 \mathrm{rad} / \mathrm{s}\) (see the drawing). The front brake is then applied for \(3.0 \mathrm{~s}\), and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s}\). Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{~m} .\) The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{k}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 300 -mile trip in a typical midsize car produces about \(1.2 \times 10^{9} \mathrm{~J}\) of energy. How fast would a \(13-\mathrm{kg}\) flywheel with a radius of \(0.30 \mathrm{~m}\) have to rotate to store this much energy? Give your answer in rev/min.

A bicycle wheel has a radius of \(0.330 \mathrm{~m}\) and a rim whose mass is \(1.20 \mathrm{~kg} .\) The wheel has 50 spokes, each with a mass of \(0.010 \mathrm{~kg}\). (a) Calculate the moment of inertia of the rim about the axle. (b) Determine the moment of inertia of any one spoke, assuming it to be a long, thin rod that can rotate about one end. (c) Find the total moment of inertia of the wheel, including the rim and all 50 spokes.

A platform is rotating at an angular speed of \(2.2 \mathrm{rad} / \mathrm{s}\). A block is resting on this platform at a distance of \(0.30 \mathrm{~m}\) from the axis. The coefficient of static friction between the block and the platform is \(0.75\). Without any external torque acting on the system, the block is moved toward the axis. Ignore the moment of inertia of the platform and determine the smallest distance from the axis at which the block can be relocated and still remain in place as the platform rotates.

A thin, rigid, uniform rod has a mass of \(2.00 \mathrm{~kg}\) and a length of \(2.00 \mathrm{~m} .\) (a) Find the moment of inertia of the rod relative to an axis that is perpendicular to the rod at one end. (b) Suppose all the mass of the rod were located at a single point. Determine the perpendicular distance of this point from the axis in part (a), such that this point particle has the same moment of inertia as the rod. This distance is called the radius of gyration of the rod.
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