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A 60.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of \(500 \mathrm{kg} \cdot \mathrm{m}^{2}\) and a radius of \(2.00 \mathrm{m} .\) The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of \(1.50 \mathrm{m} / \mathrm{s}\) relative to the Earth. (a) In what direction and with what angular speed does the turntable rotate? (b) How much work does the woman do to set herself and the turntable into motion?

Short Answer

Expert verified
a) The turntable rotates in the anticlockwise direction with an angular speed of 0.36 rad/s. b) The work done by the woman is 270 J.

Step by step solution

01

Calculate the initial Angular Momentum

Since the turntable and woman were initially at rest, the Angular Momentum \( L_{i} \) = 0.
02

Determine the woman's angular speed

The angular speed \( \omega_{w} \) of the woman relative to the turntable can be computed from her linear speed \( v \) and the radius of the turntable \( r \) using the relation \( \omega = \frac{v}{r} \). So, \( \omega_{w} = \frac{1.50 m/s}{2.00 m} = 0.75 s^{-1} \).
03

Calculate the final Angular Momentum of the woman

The Angular Momentum of the woman when she's walking is \( L_{wf} = I_{w}. \omega_{w} \), where her moment of inertia \( I_{w} = m.r^{2} = 60.0 kg . (2.00 m)^{2} = 240 kg.m^{2} \). So, \( L_{wf} = 240 kg.m^{2} . 0.75 s^{-1} = 180 kg.m^{2}/s \).
04

Calculate the final Angular Momentum of the turntable

Conservation of angular momentum dictates that the total angular momentum at any moment remains zero (since it was zero initially). Hence, the final angular momentum of the turntable would be \( L_{tf} = -L_{wf} = -180 kg.m^{2}/s \).
05

Determine the final angular speed of the turntable

We can now find the angular speed \( \omega_{t} \) of the turntable, given by \( \omega_{t} = \frac{L_{tf}}{I_{t}} \), where \( I_{t} = 500 kg.m^{2} \) is the turntable's moment of inertia. Therefore, \( \omega_{t} = \frac{-180 kg.m^{2}/s}{500 kg.m^{2}} = -0.36 s^{-1} \). Negative sign indicates that the rotation is anti-clockwise.
06

Calculate the work done by the woman

Work done is the change in kinetic energy. So, the work done by the woman, W = \( \Delta KE = \frac{1}{2} I_{w} \omega_{w}^{2} + \frac{1}{2} I_{t} \omega_{t}^{2} = \frac{1}{2} 240 kg.m^{2} (0.75 s^{-1})^2 + \frac{1}{2} 500 kg.m^{2} (-0.36 s^{-1})^{2} = 270 J \).

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Key Concepts

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

Moment of Inertia
To understand the moment of inertia, envision it as the rotational equivalent of mass in linear motion. It measures how much torque is needed for a certain angular acceleration around a pivot point. The moment of inertia depends both on the object's mass distribution relative to the axis of rotation and the axis itself. For example, a figure skater pulling their arms in reduces their moment of inertia, allowing them to spin faster with the same amount of torque. In the given exercise, the turntable has a large moment of inertia (\(500 \text{kg} \times \text{m}^{2}\)), which means it resists changes to its rotational motion, requiring significant torque to alter its angular speed.

In practical terms, the more mass an object has and the further it is distributed from its axis, the higher its moment of inertia will be. That's why the woman's location at the turntable's rim, at a maximum radius, significantly adds to the system's moment of inertia. As such, she must exert a substantial amount of force to set both herself and the turntable rotating.
Angular Speed
Angular speed describes how fast an object rotates or revolves relative to another point, in this instance, the axis of a turntable. It's the angle covered per unit of time. In our exercise, the woman's angular speed is calculated by dividing her linear speed by the radius of the circle she's traversing. Mathematically, angular speed is expressed as \( \text{angular speed} (\text{\(\omega\)}) = \frac{\text{linear speed} (\text{\(v\)})}{\text{radius} (\text{\(r\)})} \). It's measured in radians per second in the SI system.

Understanding angular speed is crucial because it enables us to comprehend the rotational dynamics of systems. For instance, knowing the angular speed aids in determining the kinetic energy of rotating bodies, as well as the stresses that materials experience under rotation, which is vital for safety and engineering.
Conservation of Angular Momentum
The principle of the conservation of angular momentum is a fundamental concept that states that if no external torque acts on a system, the total angular momentum of the system remains constant. In the case of our exercise, the system consists of the woman and the turntable. Since the system is initially at rest, its total angular momentum is zero. As the woman begins to walk and generates rotation, her movement imparts angular momentum to the system. To conserve angular momentum, the turntable rotates in the opposite direction to ensure the total remains zero. This relationship is mathematically represented as \( L_{i} = L_{f} \), where \( L_{i} \) and \( L_{f} \) are the initial and final angular momenta respectively.

This concept is essential in various real-life applications, such as understanding the behavior of spinning celestial bodies, designing mechanical systems with rotating parts, and explaining the motion of athletes in sports like diving and gymnastics.
Work and Energy in Rotation
Work and energy in the context of rotation are akin to their linear counterparts but are applied to rotational motion. The work done by a force in a rotational system is the product of torque and the angle through which it acts. The kinetic energy of a rotating body is given by \( \frac{1}{2} I \text{\(\omega^2\)} \), where \( I \) is the moment of inertia and \( \text{\(\omega\)} \) is the angular speed. In our exercise, when the woman walks and sets herself and the turntable in motion, she does work against the inertia of the system. The total work done is distributed as the kinetic energy of both the woman and the turntable in motion.

Understanding work and energy in rotation allows for an analysis of energy requirements and efficiencies of rotating systems, such as wind turbines or flywheels for energy storage. Calculating the work done and the resulting kinetic energy is essential for system design, as well as predicting the outcome of interactions in rotating objects.

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Most popular questions from this chapter

A uniform solid disk of mass \(3.00 \mathrm{kg}\) and radius \(0.200 \mathrm{m}\) rotates about a fixed axis perpendicular to its face. If the angular frequency of rotation is \(6.00 \mathrm{rad} / \mathrm{s},\) calculate the angular momentum of the disk when the axis of rotation \((a)\) passes through its center of mass and (b) passes through a point midway between the center and the rim.

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