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Chapter 10: Problem 38

A woman with mass \(50 \mathrm{~kg}\) is standing on the rim of a large horizontal disk that is rotating at \(0.80 \mathrm{rev} / \mathrm{s}\) 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
After performing the calculations in Steps 1-4, the magnitude of the total angular momentum of the woman-disk system can be obtained.

Step by step solution

01

Calculate the angular velocity

First, convert the rotating speed of the disk from revolutions per second to radians per second. Since one revolution equals \(2\pi\) radians, the angular velocity \(\omega\) equals \(0.80 \times 2\pi\) rad/s.
02

Calculate the angular momentum of the woman

The linear speed of the woman equals the product of her distance from the axis (which equals the radius of the disk) and the angular velocity of the disk. The angular momentum of the woman can then be calculated as the product of her mass, the distance from the axis, and her linear speed: \(L_{woman} = m \times r \times v = m \times r \times r \times \omega\). Substitute her mass (50 kg), radius (4.0 m), and calculated angular velocity into the equation.
03

Calculate the angular momentum of the disk

The angular momentum of the disk equals the product of its moment of inertia and angular velocity. The moment of inertia for a disk rotating about an axis through its center equals \(\frac{1}{2} M r^{2}\), where \(M\) is the mass of the disk and \(r\) is its radius. The angular momentum \(L_{disk}\) then equals \(I \times \omega = 0.5 \times M \times r^2 \times \omega\). Substitute the mass of the disk (110 kg), its radius (4.0 m), and the calculated angular velocity into the equation.
04

Calculate the total angular momentum

Add the angular momentum of the woman and the disk to get the total angular momentum of the woman-disk system: \(L_{total} = L_{woman} + L_{disk}\).

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Key Concepts

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

Angular Velocity
Angular velocity is a vector quantity which denotes the rate of change of angular position of an object rotating about an axis. In simpler terms, it describes how fast an object spins or rotates. Like linear speed, which tells us how fast something moves in a straight line, angular velocity tells us how quickly something is turning.

To understand angular velocity, imagine you're on a merry-go-round; the speed at which you're turning around the center is related to the angular velocity. The standard unit of angular velocity is radians per second (rad/s). One revolution per second is equivalent to an angular velocity of \(2\pi\) rad/s, as \(2\pi\) radians represents a full circle. This concept is crucial in solving rotational motion problems, as seen with the rotating disk in the woman-disk system exercise.
Moment of Inertia
The moment of inertia, often symbolized by \(I\), is a measure of an object's resistance to changes to its rotation. It depends on the mass distribution relative to the rotation axis. Just as mass influences how much force is needed to accelerate an object in linear motion, the moment of inertia plays an analogous role for rotating objects.

For different shapes and mass distributions, there are specific formulas to calculate moment of inertia. For a solid disk rotating around its center, the formula is \(\frac{1}{2} M r^{2}\), where \(M\) is the mass and \(r\) is the radius. Moment of inertia is significant because it directly affects the angular momentum, where a larger moment of inertia means more torque (rotational force) is required to change its angular velocity.
Rotational Kinematics
Rotational kinematics is the study of the motion of objects that rotate or spin. Its principles are analogous to translational kinematics, where distances, displacements, velocities, and accelerations are replaced with angular distances, angular displacements, angular velocities, and angular accelerations.

Understanding rotational kinematics allows us to predict the future position, velocity, and acceleration of a rotating object given initial conditions. The equations take a similar form to those of linear kinematics but includes rotational concepts like angular displacement (symbolized by \(\theta\)) and angular acceleration (symbolized by \(\alpha\)). Crucially, the kinematic equations for rotation relate the angular velocity, angular acceleration, and time to determine various aspects of rotational motion, such as how long it takes for an object to stop spinning if a certain torque is applied.

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Most popular questions from this chapter

The moment of inertia of the empty turntable is \(1.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\). With a constant torque of \(2.5 \mathrm{~N} \cdot \mathrm{m},\) the turntable-person system takes \(3.0 \mathrm{~s}\) to spin from rest to an angular speed of \(1.0 \mathrm{rad} / \mathrm{s} .\) What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. (a) \(2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (b) \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (c) \(7.5 \mathrm{~kg} \cdot \mathrm{m}^{2} ;\) (d) \(9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\).

Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass \(5.00 \mathrm{~kg}\) and radius \(0.120 \mathrm{~m} .\) For each the translational speed of the center of mass is \(4.00 \mathrm{~m} / \mathrm{s}\). Sphere \(A\) is a uniform solid sphere and sphere \(B\) is a thin-walled, hollow sphere. How much work, in joules, must be done on each sphere to bring it to rest? For which sphere is a greater magnitude of work required? Explain. (The spheres continue to roll without slipping as they slow down.

You are designing a system for moving aluminum cylinders from the ground to a loading dock. You use a sturdy wooden ramp that is \(6.00 \mathrm{~m}\) long and inclined at \(37.0^{\circ}\) above the horizontal. Each cylinder is fitted with a light, frictionless yoke through its center, and a light (but strong) rope is attached to the yoke. Each cylinder is uniform and has mass 460 \(\mathrm{kg}\) and radius \(0.300 \mathrm{~m} .\) The cylinders are pulled up the ramp by applying a constant force \(\overrightarrow{\boldsymbol{F}}\) to the free end of the rope. \(\overrightarrow{\boldsymbol{F}}\) is parallel to the surface of the ramp and exerts no torque on the cylinder. The coefficient of static friction between the ramp surface and the cylinder is \(0.120 .\) (a) What is the largest magnitude \(\overrightarrow{\boldsymbol{F}}\) can have so that the cylinder still rolls without slipping as it moves up the ramp? (b) If the cylinder starts from rest at the bottom of the ramp and rolls without slipping as it moves up the ramp, what is the shortest time it can take the cylinder to reach the top of the ramp?

A size-5 soccer ball of diameter \(22.6 \mathrm{~cm}\) and mass \(426 \mathrm{~g}\)rolls up a hill without slipping, reaching a maximum height of \(5.00 \mathrm{~m}\) above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it have then? Neglect rolling friction and assume the system's total mechanical energy is conserved.

A machine part has the shape of a solid uniform sphere of mass \(225 \mathrm{~g}\) and diameter \(3.00 \mathrm{~cm}\). It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of \(0.0200 \mathrm{~N}\) at that point. (a) Find its angular acceleration. (b) How long will it take to decrease its rotational speed by \(22.5 \mathrm{rad} / \mathrm{s} ?\)
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