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Chapter 10: Problem 36

A woman with mass 50 \(\mathrm{kg}\) is standing on the rim of a large disk that is rotating at 0.50 \(\mathrm{rev} / \mathrm{s}\) about an axis through its center. The disk has mass 110 \(\mathrm{kg}\) and radius 4.0 \(\mathrm{m} .\) Calculate the magnitude of the total angular momentum of the woman-disk system. (Assume that you can treat the woman as a point.)

Short Answer

Expert verified
The total angular momentum of the woman-disk system is approximately 5275 kg路m虏/s.

Step by step solution

01

Understand the Problem

We have a woman standing on the rim of a rotating disk. We need to calculate the total angular momentum of the system composed of the woman and the disk, about the axis of rotation.
02

Use Conservation of Angular Momentum

The total angular momentum of the system is the sum of the angular momentum of the woman and the disk itself. We need to calculate these values separately and sum them.
03

Calculate Angular Momentum of the Disk

The angular momentum of a solid disk rotating about its central axis is given by the formula \( L = I \omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. The moment of inertia \( I \) for a disk is \( \frac{1}{2}mR^2 \). Substitute \( m = 110\,\mathrm{kg} \), \( R = 4.0\, \mathrm{m} \), and \( \omega = 0.50 \times 2\pi\,\mathrm{rad/s} \).
04

Substitute Values for Disk

The moment of inertia \( I \) for the disk is \( \frac{1}{2} \times 110\,\mathrm{kg} \times (4.0\,\mathrm{m})^2 = 880\, \mathrm{kg}\cdot\mathrm{m}^2\). The angular velocity \( \omega = 0.50 \times 2\pi\,\mathrm{rad/s} \approx 3.14\, \mathrm{rad/s} \). Thus, the angular momentum of the disk \( L_{disk} = 880\, \mathrm{kg}\cdot\mathrm{m}^2 \times 3.14\, \mathrm{rad/s} \approx 2763\, \mathrm{kg}\cdot\mathrm{m}^2/s \).
05

Calculate Angular Momentum of the Woman

Treat the woman as a point mass. Her angular momentum is given by \( L = mvr \), where \( v = r\omega \) is her linear velocity. Substitute \( m = 50\, \mathrm{kg}\), \( r = 4.0\, \mathrm{m} \), and \( \omega = 3.14\, \mathrm{rad/s} \).
06

Substitute Values for Woman

The linear velocity \( v = 4.0\, \mathrm{m} \times 3.14\, \mathrm{rad/s} = 12.56\, \mathrm{m/s} \). Therefore, the angular momentum \( L_{woman} = 50\, \mathrm{kg} \times 12.56\, \mathrm{m/s} \times 4.0\, \mathrm{m} = 2512\, \mathrm{kg}\cdot\mathrm{m}^2/s \).
07

Calculate Total Angular Momentum

The total angular momentum is the sum of the angular momenta of the disk and the woman: \( L_{total} = L_{disk} + L_{woman} = 2763\, \mathrm{kg}\cdot\mathrm{m}^2/s + 2512\, \mathrm{kg}\cdot\mathrm{m}^2/s = 5275\, \mathrm{kg}\cdot\mathrm{m}^2/s \).

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Key Concepts

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

Moment of Inertia
The concept of moment of inertia is crucial in understanding rotational dynamics. It acts as a rotational analog to mass in linear motion, indicating how difficult it is to change an object's rotational state. The moment of inertia depends on both the mass of an object and the distribution of that mass relative to the axis of rotation. For a solid rotating disk, the moment of inertia is calculated using the formula \[ I = \frac{1}{2} m R^2 \]where \( m \) is the mass of the disk and \( R \) is its radius. This formula shows that the inertia depends not only on the disk's mass but quadratically on the radius, meaning that increasing the radius significantly impacts the moment of inertia.
  • Mass and radius are key factors.
  • A larger radius leads to more rotational inertia.
Understanding these properties helps predict how any rotating body will respond to torques and accelerations.
Rotational Dynamics
Rotational dynamics involves the motion of objects when they rotate or spin. Key principles include angular velocity, torque, and angular momentum. These concepts mirror those in linear dynamics, such as velocity, force, and momentum. Angular velocity \( \omega \) represents how fast something rotates, typically in radians per second. The change in angular velocity occurs through torque, the rotational equivalent of force. To maintain equilibrium or analyze motion in instances like our exercise with the woman and the disk, understanding these dynamics is essential. The angular momentum \( L \) of a rotating object is given by the product of its moment of inertia and angular velocity:\[ L = I\omega \]
  • Torque changes angular velocity.
  • Angular momentum is conserved.
These principles allow us to calculate and predict the behavior of rotating systems, helping explain the disk-woman setup.
Conservation of Angular Momentum
The conservation of angular momentum is a fundamental principle in physics stating that if no external torque acts on a system, the total angular momentum remains constant. This principle is akin to the conservation of linear momentum in straight-line motion. It applies to various scenarios, from simple rotating wheels to complex celestial systems.In the exercise, the idea is pivotal for computing the total angular momentum of the woman-disk system. Since neither the woman nor the disk exerts external torque on the system, the angular momentum is conserved.
  • One rotating body can't change the total \( L \) without external forces.
  • Helps confirm correctness of calculations.
This conservation law is crucial in understanding how isolated systems behave, drawing links between the initial and final states of motion.

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Most popular questions from this chapter

A lawn roller in the form of a thin-walled, hollow cylinder with mass \(M\) is pulled horizontally with a constant horizontal force \(F\) applied by a handle attached to the axle. If it rolls without slipping, find the acceleration and the friction force.

A \(2.20-\) kg hoop 1.20 \(\mathrm{m}\) in diameter is rolling to the right without slipping on a horizontal floor at a steady 3.00 \(\mathrm{rad} / \mathrm{s}\) . (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

A solid disk is rolling without slipping on a level surface at a constant speed of 3.60 \(\mathrm{m} / \mathrm{s}\) (a) If the disk rolls up a \(30.0^{\circ}\) ramp, how far along the ramp will it move before it stops? (b) Explain why your answer in part (a) does not depend on either the mass or the radius of the disk.

Sedna. In November \(2003,\) the now-most-distant-known object in the solar system was discovered by observation with a telescope on Mt. Palomar. This object, known as Sedna, is approximately 1700 \(\mathrm{km}\) in diameter, takes about \(10,500\) years to orbit our sun, and reaches a maximum speed of 4.64 \(\mathrm{km} / \mathrm{s}\) . Calculations of its complete path, based on several measurements of its position, indicate that its orbit is highly elliptical, varying from 76 \(\mathrm{AU}\) to 942 \(\mathrm{AU}\) in its distance from the sun, where \(\mathrm{AU}\) is the astronomical unit, which is the average distance of the earth from the sun \(\left(1.50 \times 10^{8} \mathrm{km}\right)\) . (a) What is Sedna's minimum speed? (b) At what points in its orbit do its maximum and minimum speeds occur? (c) What is the ratio of Sedna's maximum kinetic energy to its minimum kinetic energy?

A \(50.0\)-kg grindstone is a solid disk 0.520 \(\mathrm{m}\) in diameter. You press an ax down on the rim with a normal force of 160 \(\mathrm{N}\) (Fig. P10.57). The coefficient of kinetion between the blade and the stone is 0.60 , and there is a constant friction torque of 6.50 \(\mathrm{N} \cdot \mathrm{m}\) between the axle of the stone and its bearings. (a) How much force must be applied tangentially at the end of a crank handle 0.500 \(\mathrm{m}\) long to bring the stone from rest to 120 \(\mathrm{rev} / \mathrm{min}\) in 9.00 \(\mathrm{s} ?\) (b) After the grindstone attains an angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) , what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) ? (c) How much time does it take the grindstone to come from 120 \(\mathrm{rev} / \mathrm{min}\) to rest if it is acted on by the axle friction alone?
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