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Chapter 10: Problem 38

A potter's wheel-a thick stone disk of radius \(0.500 \mathrm{m}\) and mass \(100 \mathrm{kg}\) - is freely rotating at \(50.0 \mathrm{rev} / \mathrm{min} .\) The potter can stop the wheel in 6.00 s by pressing a wet rag against the rim and exerting a radially inward force of \(70.0 \mathrm{N}\) Find the effective coefficient of kinetic friction between wheel and rag.

Short Answer

Expert verified
The effective coefficient of kinetic friction between the wheel and the rag is 0.157.

Step by step solution

01

Conversion of units

Firstly, the angular speed of the wheel needs to be converted from revolutions per minute (rev/min) to a standard unit of angular speed, radian per second (rad/s). One revolution is equivalent to \(2\pi\) radians, and one minute is equivalent to 60 seconds. Hence, \(50.0 \, rev/min\) equals \(50.0 * 2\pi / 60 = 5.23599 \, rad/s\).
02

Calculate the initial angular momentum

Angular momentum \(L\) is the product of moment of inertia \(I\) and angular velocity \(\omega\). The moment of inertia of a disk is \(I = 0.5 * m * r^2\), where \(m = 100\, kg\) is the mass of the disk and \(r = 0.5\, m\) is the radius. So, the initial angular momentum \(L_i\) is \(0.5 * m * r^2 * \omega = 0.5 * 100 * 0.5^2 * 5.23599 = 32.98 \, kgm^2/s\).
03

Calculate the final angular momentum and angular deceleration

Given that the wheel stops, the final angular momentum \(L_f\) is zero. Since the wheel slows down and stops in 6 seconds, the angular deceleration \(\alpha\) can be found by \(\alpha = (\omega_f - \omega_i) / t = (0 - 5.23599) / 6 = -0.873 \, rad/s^2\).
04

Calculate the Torque

The torque \(\tau\) is obtained by multiplying the moment of inertia \(I\) by the angular deceleration \(\alpha\). Hence, \(\tau = I * \alpha = 0.5 * m * r^2 * \alpha = 0.5 * 100 * 0.5^2 * -0.873 = -5.482 \, Nm\). The minus sign indicates that the direction of torque is opposite to the initial direction of rotation, which makes the wheel decelerate.
05

Calculate the friction force

The friction force acting on the wheel's rim can be obtained by dividing the torque by the radius of the wheel: \(f = \tau / r = -5.482 / 0.5 = -10.964 \, N\). The negative sign suggests the friction force acts opposite to the wheel鈥檚 initial rotational direction.
06

Calculate the normal force

The normal force is the radial inward force exerted by the potter which is 70.0 N.
07

Find the coefficient of kinetic friction

Lastly, the effective coefficient of kinetic friction \(\mu_k\) is obtained by \(\mu_k = f / N = -10.964 / 70.0 = -0.157\). The negative sign is ignored when reporting the friction coefficient as it's a scalar value and the direction has been accounted for already.

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Key Concepts

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

Angular Momentum
Angular momentum is a key concept in understanding rotational motion. It's similar to linear momentum but instead applies to objects that are spinning or rotating. It's represented by the symbol L, and is calculated as the product of an object's moment of inertia (I) and its angular velocity (蝇). Mathematically, L = I * 蝇.

In our potter's wheel scenario, the wheel's initial angular momentum is substantial due to its mass and speed of rotation. As the potter applies a force, they generate a torque that changes the angular momentum over time, leading to the wheel's eventual stop. Remember, the principle of conservation of angular momentum tells us that in the absence of an external torque, the total angular momentum of a system remains constant, which we can observe in the wheel's free rotation before the rag is applied.
Moment of Inertia
The moment of inertia is the rotational equivalent of mass for linear motion and quantifies an object's resistance to changes in its rotation rate. It is highly dependent on the object's mass distribution relative to the axis of rotation. For a disk like the potter's wheel, it is represented by the formula I = 0.5 * m * r2, where m is the mass and r is the radius of the disk.

The larger the moment of inertia, the harder it is to change the object鈥檚 rotational speed. Every object has its unique moment of inertia calculated based on shape and mass distribution, which tells us the amount of torque required to achieve a certain angular acceleration.
Angular Deceleration
Angular deceleration, denoted by 伪, measures how quickly a rotating object slows down. It's the rate of change of angular velocity with time. To find the angular deceleration, you take the change in angular velocity and divide it by the time it takes for that change to occur: 伪 = (蝇f - 蝇i) / t.

In the potter's wheel case, as the potter applies the rag to the wheel, the angular deceleration tells us how rapidly the wheel comes to a halt. The deceleration is negative because the wheel is slowing down; it's the opposite process of acceleration in which something speeds up.
Torque
Torque is the measure of the force that can cause an object to rotate around an axis. Just as a force is a push or a pull, a torque can be thought of as a twist to an object. Torque (蟿) = radius (r) * force (f), and in this case, it鈥檚 the force applied by the wet rag at a distance from the wheel鈥檚 center.

The direction of torque is important and is determined by the right-hand rule. In our exercise, the torque created by the potter's action is negative, indicating it acts in the direction opposite the wheel's rotation. The greater the torque applied to an object, the greater the object's rotation rate will change 鈥� which goes hand in hand with the concept of angular deceleration.

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Most popular questions from this chapter

A videotape cassette contains two spools, each of radius \(r_{\mathrm{g}}\) on which the tape is wound. As the tape unwinds from the first spool, it winds around the second spool. The tape moves at constant linear speed \(v\) past the heads between the spools. When all the tape is on the first spool, the tape has an outer radius \(r_{t}\). Let \(r\) represent the outer radius of the tape on the first spool at any instant while the tape is being played. (a) Show that at any instant the angular speeds of the two spools are $$\omega_{1}=v / r \quad \text { and } \quad \omega_{2}=v /\left(r_{s}^{2}+r_{t}^{2}-r^{2}\right)^{1 / 2}$$ (b) Show that these expressions predict the correct maximum and minimum values for the angular speeds of the two spools.

The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for \(8.00 \mathrm{s}\), at which time it is turning at 5.00 rev/s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub smoothly slows to rest in \(12.0 \mathrm{s}\) Through how many revolutions does the tub turn while it is in motion?

The combination of an applied force and a friction force produces a constant total torque of \(36.0 \mathrm{N} \cdot \mathrm{m}\) on a wheel rotating about a fixed axis. The applied force acts for \(6.00 \mathrm{s} .\) During this time the angular speed of the wheel increases from 0 to 10.0 rad/s. The applied force is then removed, and the wheel comes to rest in 60.0 s. Find (a) the moment of inertia of the wheel, (b) the magnitude of the frictional torque, and (c) the total number of revolutions of the wheel.

An electric motor rotating a grinding wheel at \(100 \mathrm{rev} / \mathrm{min}\) is switched off. With constant negative angular acceleration of magnitude \(2.00 \mathrm{rad} / \mathrm{s}^{2},\) (a) how long does it take the wheel to stop? (b) Through how many radians does it turn while it is slowing down?

A thin rod of mass \(0.630 \mathrm{kg}\) and length \(1.24 \mathrm{m}\) is at rest, hanging vertically from a strong fixed hinge at its top end. Suddenly a horizontal impulsive force \((14.7 \mathrm{i}) \mathrm{N}\) is applied to it. (a) Suppose the force acts at the bottom end of the rod. Find the acceleration of its center of mass and the horizontal force the hinge exerts. (b) Suppose the force acts at the midpoint of the rod. Find the acceleration of this point and the horizontal hinge reaction. (c) Where can the impulse be applied so that the hinge will exert no horizontal force? This point is called the center of percussion.
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