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.