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(III) Suppose a 65 -kg person stands at the edge of a 6.5 -m dia- meter merry-go-round turntable that is mounted on frictionless bearings and has a moment of incrtia of 1850 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) . The tumtable is at rest initially, but when the person begins rumning at a speed of 3.8 \(\mathrm{m} / \mathrm{s}\) (with respect to the tumtablc) around its edge, the turntable begins to rotate in the opposite direction. Cakulate the angular velocity of the tumtable.

Short Answer

Expert verified
The angular velocity of the turntable is approximately 0.436 rad/s.

Step by step solution

01

Understand the Problem

We have a system containing a person and a merry-go-round. Initially, both are at rest. When the person starts running, conservation of angular momentum dictates that the merry-go-round will rotate in the opposite direction. We need to calculate its angular velocity.
02

Set Up Conservation of Angular Momentum

Angular momentum is conserved because there is no external torque acting on the system. We write the conservation equation as: \[ L_{initial} = L_{final} \]Since the system is initially at rest, \( L_{initial} = 0 \).
03

Express Final Angular Momentum

The final angular momentum is the sum of the angular momentum of the person and the angular momentum of the turntable. Let the mass of the person be \( m = 65 \text{ kg} \), velocity \( v = 3.8 \text{ m/s} \), moment of inertia of the turntable \( I = 1850 \text{ kg m}^2 \), and radius \( r = 3.25 \text{ m} \). The final equation is:\[ L_{person} + L_{turntable} = 0 \] \[ mvr - I\omega = 0 \]
04

Solve for Angular Velocity

Rearrange the equation to solve for \( \omega \), the angular velocity of the turntable:\[ mvr = I\omega \]Substitute the known values:\[ (65 \text{ kg})(3.8 \text{ m/s})(3.25 \text{ m}) = 1850 \text{ kg m}^2 \times \omega \]\[ 806.25 \text{ kg m}^2/s = 1850 \text{ kg m}^2 \times \omega \]\[ \omega = \frac{806.25}{1850} \approx 0.436 \text{ rad/s} \]
05

Interpret the Result

The negative sign indicates the direction of rotation is opposite to the direction of the person's movement. However, we are usually interested in the magnitude of the angular velocity, so we state it as \( 0.436 \text{ rad/s} \).

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Key Concepts

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

Conservation of Angular Momentum
Conservation of angular momentum is a fundamental principle in physics. It states that in a closed system with no external torques, the total angular momentum remains constant. This principle is crucial in problems involving rotating systems, like our merry-go-round and the person. When the person begins to run, they exert an internal force, causing a change in their own angular momentum.

To conserve angular momentum, the merry-go-round must rotate in the opposite direction. The system's initial angular momentum is zero because both the person and the turntable start at rest. Thus, their final angular momentum must also be zero to comply with conservation laws. This interplay ensures that any increase in the person's angular momentum due to their motion is offset by a corresponding change in the turntable's rotation.
Moment of Inertia
Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It's analogous to mass in linear motion but applies to rotational dynamics. The moment of inertia depends on how mass is distributed in relation to the axis of rotation. For complex shapes and distributions, calculating the moment of inertia can be more challenging.

In the case of the merry-go-round, the moment of inertia is given as 1850 \( ext{kg} \cdot ext{m}^2 \). This value signifies how much effort it takes to change its rotational speed. When solving the problem, understanding that the moment of inertia remains constant allows us to use this value to determine the angular velocity once the person begins to run.
  • High moment of inertia: Requires more torque to change rotational speed.
  • Low moment of inertia: Easier to spin or stop.
Angular Velocity
Angular velocity describes how fast an object rotates around a specific point or axis. It is measured in radians per second (rad/s) and indicates the rate of rotation. After the person runs around the edge of the turntable, the surface begins to spin in the opposite direction to maintain angular momentum.

Angular velocity was calculated in the example by rearranging the conservation of angular momentum equation: \[ mvr = I\omega \]and plugging in the known values to solve for \( \omega \), the angular velocity of the turntable. This concept is key in analyzing rotational motions, whether in physics problems or real-world applications like gear systems and rotating machinery.
Physics Problem Solving
Approaching physics problems methodically can greatly assist in understanding and solving them accurately. Breaking down the given problem into steps streamlines the thought process and helps identify what principles and formulas are applicable. Let's look at the approach used in this scenario:

  • Identify the known quantities (mass, velocity, and moment of inertia).
  • Apply the conservation of angular momentum, recognizing that the initial state is zero.
  • Write down the final expressions for angular momentum for individual system components.
  • Rearrange and solve the equation to find the desired quantity, such as angular velocity.
By structuring problem solving with clear steps, students can apply a similar strategy across various physics challenges. Understanding fundamental physics concepts and practicing their application helps in mastering these methodologies, ensuring success in problem-solving.

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Most popular questions from this chapter

Suppose our Sun eventually collapses into a white dwarf, losing about half its mass in the process, and winding up with a radius \(1.0 \%\) of its existing radius. Assuming the lost mass carries away no angular momentum, what would the Sun's new rotation rate be? (Take the Sun's current period to be about 30 days.) What would be its final kinetic energy in terms of its initial kinetic energy of today?

(II) Two identical particles have equal but opposite momenta, \(\vec{\mathbf{p}}\) and \(-\vec{\mathbf{p}}\) , but they are not traveling along the same line. Show that the total angular momentum of this system does not depend on the choice of origin.

A 70 -kg person stands on a tiny rotating platform with arms outstretched. (a) Estimate the moment of inertia of the person using the following approximations: the body (including head and legs) is a 60 -kg cylinder, 12 \(\mathrm{cm}\) in radius and 1.70 \(\mathrm{m}\) high; and each arm is a \(5.0-\mathrm{kg}\) thin rod, 60 \(\mathrm{cm}\) long, attached to the cylinder. (b) Using the same approximations, estimate the moment of inertia when the arms are at the person's sides. (c) If one rotation takes 1.5 \(\mathrm{s}\) when the person's arms are outstretched, what is the time for each rotation with arms at the sides? Ignore the moment of inertia of the lightweight platform. (d) Determine the change in kinetic energy when the arms are lifted from the sides to the horizontal position, (e) From your answer to part \((d),\) would you expect it to be harder or easier to lift your arms when rotating or when at rest?

(II) A nonrotating cylindrical disk of moment of inertia \(I\) is dropped onto an identical disk rotating at angular specd \omega. Assuming no external torques, what is the final common angular spced of the two disks?

A spherical asteroid with radius \(r=123 \mathrm{m}\) and mass \(M=2.25 \times 10^{10} \mathrm{kg}\) rotates about an axis at four revolutions per day. A "tug" spaceship attaches itself to the pole (as defined by the axis of rotation) and fires its engine, applying a force \(F \quad\) tangentially to the asteroid's surface as shown in Fig. \(44 .\) If \(F=265 \mathrm{N},\) how long will it take the tug to rotate the asteroid's axis of rotation through an angle of \(10.0^{a}\) by this method?

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