Torque and angular acceleration of a grindstone

In summary, we have a grindstone with a diameter of 0.550 m and a mass of 50.0 kg rotating at 840 rev/min. When pressed with a normal force of 160 N, the grindstone comes to rest in 7.60 s. To find the coefficient of friction between the ax and the grindstone, we can use Newton's second law for rotation. The friction force is related to the normal force, and we can use the equation torque = r x (Fnmewk) = I x alpha to find the coefficient of friction. We can convert ω to units of radians/second and solve for mewk, which should not be larger than 1.
  • #1
BoldKnight399
79
0
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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  • #2
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?
 
  • #3
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?
 
  • #4
Yes, the friction force is the force that produces the torque.
 
  • #5
so then torque=r X (fn*mewk)=I*alpha
but then how do I relate omega to alpha? is it alpha=domega/dt?
 
  • #6
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.
 
  • #7
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?
 
  • #8
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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