Torque/ find the coefficient of friction

  • Thread starter Mdhiggenz
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  • #1
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Homework Statement



A grindstone in the shape of a solid disk with diameter 0.520m and a mass of 50.0kg is rotating at 800rev/min . You press an ax against the rim with a normal force of 160N, and the grindstone comes to rest in 8.50s .
Find the coefficient of kinetic friction between the ax and the grindstone.


Homework Equations





The Attempt at a Solution



The concept of torque has been giving me alot of problems, mostly UNDERSTANDING what the equation means, and when to use them.

So here is how I started out,

I=1/2MR^2

850rev/min= 85pi rad/3s

So my logic is I have the inertia, and I have enough information to solve for acceleration which means I can find the torque.

I solve for the acceleration

wz-woZ/t=-11.9rad/s^2

Here is where I get a bit confused, where I was going with this problem is I was going to solve for the torque using Ʃτz= Iαz

Then when I aquired the torque I was going to use

ƩFext=Macm

to find the coefficient of friction.

The solution manual sets Ʃτz= μkNR=Iαz, and solves for μk.

How do we know when to stay at Ʃτz=Iaz or when we have to switch over to ƩFext=Macm?

Thank you

Higgenz
 

Answers and Replies

  • #2
BruceW
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In the problem statement, you say it is originally rotating at 800rev/min, but in the attempt at the solution, you are using 850rev/min... was this a typing error?

About the torque: the torque is due to a force at the rim of the grindstone. There is an equation to calculate the torque due to a force at a point, but the equation is not ƩFext=Macm
 
  • #3
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Typing error is is originally 850 rev/min and 7.50s apologies for that.
 
  • #4
BruceW
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good, I get the same answer of 11.9rad/s^2 for the angular acceleration.

So now you can go to the next step, "I was going to solve for the torque using Ʃτz= Iαz" And then after that, what equation relates torque to the force and the position that force is applied at?
 
  • #5
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the lever arm R?
 
  • #6
BruceW
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yep, and generally it also involves the angle between the force and the lever arm, but this case happens to be simple, because of what?

Edit: and so, what is the equation between torque and applied force? You've already got a big clue, since you've seen the solution manual
 
  • #7
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Would it be because we are dealing with angular, and not a mix between angular and linear motion, and the equation bewteen torque and applied force would be torque= FR
 
  • #8
BruceW
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You are right that it is simpler because it's purely angular motion. But I was talking about the general equation between force and torque:
[tex]torque = RF sin(\theta)[/tex]
And you're right that we can use the simpler version torque=FR, but why are we allowed to say this? (I'm just trying to make sure you've thought through each step properly).

And then what is the next step?
 
  • #9
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That is where I get a big confused there are so many equations that can give you the same exact thing.

Torque=FL
Torque=RF

I also dont understand why they use sin(theta) on some problems it would require you to find another angle, when it would be easier to use cos theta instead.
 
  • #10
BruceW
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I'm not sure what you mean about finding another angle... But in this problem, we have the equation torque = RF sin(theta), or you might know it better as Force times perpendicular distance. Anyway, the question tells us that there is a frictional force at the rim. So what is the direction of the frictional force, and therefore what is theta? (and so what is the torque?)
 

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