Why Is the Coefficient of Kinetic Friction Calculation Off by a Factor of 2?

[Leeoku]

Homework Statement

A potter's wheel—a thick stone disk with a radius of 0.450 m and a mass of 128 kg—is freely rotating at 52.0 rev/min. The potter can stop the wheel in 5.95 s by pressing a wet rag against the rim and exerting a radially inward force of 82.0 N. Calculate the effective coefficient of kinetic friction between the wheel and the rag.

Homework Equations

f_k = mu_k N
Torque = I*alpha (angular acceleration)= F*D

The Attempt at a Solution

I got most of it except i don't know why it is off by a factor of 2
Convert revs/min to angular accelearation
alpha = 52*2pi/60/5.95
=.915 s
Using Torque = I*alpha = F*d
F = I*alpha/d
= m(r^2)alpha/d
= 128kg*(.45m^2)*.915rads/s^2
--------------------------------------
.45 m
= 52.7 N
f_k = mu_k N
52.7 = 82mu_k
.64 = u

Correct answer is 0.321

 

[Filip Larsen]

Welcome to PF!

Check your moment of inertia.

 

[Leeoku]

oh there it is, i just noticed from my textbook I for disk = mr^2/2. thx!

Random side question, is it important to know centre of mass for moment of inertia questions? My prof didn't teach it yet and my test is 2moro and I'm scared of torque questions that might include it. What scenarios would require Centre of mass in torques

 

[Filip Larsen]

In scenarios that involves simple rotation about a fixed axis where only torques are interesting, the center of mass can usually be ignored. The center of mass becomes important when you have free rotation or need to know the bearing forces resulting from a rotation axis that do not pass through the center of mass. And for the general combined translational and rotational dynamics of rigid bodies, the center of mass is certainly a very important concept.

If you haven't been taught about center of mass yet it seems strange that your professor would give a test that requires such knowledge.

 

[rwisz]

.

Hello,

Thank you for providing the problem and your solution attempt. I see that you have correctly used the equation for torque, which is Torque = I*alpha = F*d. However, there are a few mistakes in your calculations that have led to the incorrect answer.

Firstly, when converting revolutions per minute to radians per second, you have multiplied by 2pi/60 but have not multiplied by the number of revolutions (52). This should give you an angular acceleration of 5.43 rad/s^2.

Secondly, the mass and radius of the wheel should be used to calculate the moment of inertia (I), not just the radius. The moment of inertia for a solid disk is given by I = (1/2)*m*r^2. This gives you a moment of inertia of 5.76 kg*m^2.

Using these corrected values, the force (F) needed to stop the wheel can be calculated as:

F = I*alpha/d
= (5.76 kg*m^2)*(5.43 rad/s^2)/0.45 m
= 69.3 N

Finally, using the equation for kinetic friction, f_k = mu_k*N, the coefficient of kinetic friction can be calculated as:

mu_k = f_k/N
= 69.3 N/82 N
= 0.844

This is the effective coefficient of kinetic friction between the wheel and the rag. However, since this is a rotating system, the effective coefficient of kinetic friction is actually half of this value. This is because the force is applied at the rim of the wheel, but the frictional force acts at the center of the wheel, which is half the radius away. Therefore, the correct answer is:

mu_k = 0.844/2 = 0.422

I hope this helps clarify the solution and the factor of 2 difference. Keep up the good work in your studies!