Torque and angular acceleration of a grindstone

AI Thread Summary
The discussion focuses on calculating the coefficient of friction between an ax and a grindstone, which is a solid disk with specific dimensions and mass. Participants suggest using Newton's second law for rotation and relate the normal force to the friction force, emphasizing that torque is produced by the friction force. There is confusion regarding the conversion of angular velocity from revolutions per minute to radians per second, which is necessary for accurate calculations. One participant calculates an implausibly high coefficient of friction, indicating a potential error in their calculations. The conversation highlights the importance of correctly applying physics principles and unit conversions in solving the problem.
BoldKnight399
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A grindstone in the shape of a solid disk with diameter 0.550 m and a mass of m = 50.0 kg is rotating at omega = 840 rev/min. You press an ax against the rim with a normal force of F = 160 N , and the grindstone comes to rest in 7.60 s. Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

To be honest, I want to draw a force diagram, but the thing is that the force then pushes into the grindstone. And then I was thinking about using Wtot-Wfriction=Change in Kinetic energy. But the thing is that I don't know how to use the Fnormal. If anyone can point me in the right direction, that would be great.
 
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BoldKnight399 said:
And then I was thinking about using Wtot-Wfriction=Change in Kinetic energy.
Better to use Newton's 2nd law for rotation.
But the thing is that I don't know how to use the Fnormal.
Hint: How does Fnormal relate to the friction force?
 
second law for rotations? torque=rXF? and Oh, fk=fnmewk. but I still don't know where to fit it into the equation/ or is that my F?
 
Yes, the friction force is the force that produces the torque.
 
so then torque=r X (fn*mewk)=I*alpha
but then how do I relate omega to alpha? is it alpha=domega/dt?
 
BoldKnight399 said:
so then torque=r X (fn*mewk)=I*alpha
Good.
but then how do I relate omega to alpha? is it alpha=domega/dt?
Yes. α = Δω/Δt. But convert ω to units of radians/second.
 
So i did that using my numbers and I get that the answer is 29.83 which makes absolutely no sense because mew cannot be bigger than 1. Did I do somethign wrong in the calculations: I got that my equations were:
r*Fnmewk=1/2mr^2*(w*2pi/t)
so
mewk=(1/2mr^2*(w*2pi/t))/(r*F)
so mew k=29 which is wrong
so did I just plug it into my calculator wrong or what?
 
BoldKnight399 said:
Did I do somethign wrong in the calculations: I got that my equations were:
r*Fnmewk=1/2mr^2*(w*2pi/t)
Note that you are given ω in units of rev/minute. How can you convert to radians/second?
 
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