Solid disk frictional torque magnitude

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1. A solid disk of radius R and mass M is initially spinning clockwise with an angular velocity magnitude ω. It is placed against a flat, horizontal surface with a coefficient of kinetic friction of μ.

Find the frictional torque magnitude on the disk, and how long it will take the disk to come to rest.


Homework Equations



I know that for finding how long it will take to stop I need ω=ω0+αt and to find α I use T=Iα.
However, I'm not sure how to find T for the first part.
I also think that I might need the inertia of a disk, so that's I=½MR2

3. The Attempt at a Solution


Alright so as I said I pretty much know how to find the time needed for it to come to a stop, however how in the world do I find the torque? All of my equations for finding torque involve F, like T=rFsinθ, but the problem has no F, is there a different equation I should be using or something?

I had originally tried using the T=rFsinθ and had θ=90, and I made the Force be the Fk, but apparently that's wrong cause I got marked off for it. Please I need some clear guidance.
 
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Answers and Replies

  • #2
haruspex
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It is not clear, but I assume the disk is laid onto a horizontal surface, so the normal force is the weight of the disk.
You will need to assume the normal force is distributed evenly over the surface of the disk.
Consider an element radius r, width dr, length ##rd\theta##. What is the frictional force on that? What torque does it exert about the disk's centre?
 
  • #3
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It is not clear, but I assume the disk is laid onto a horizontal surface, so the normal force is the weight of the disk.
You will need to assume the normal force is distributed evenly over the surface of the disk.
Consider an element radius r, width dr, length ##rd\theta##. What is the frictional force on that? What torque does it exert about the disk's centre?

Alright so the Force becomes F=Mg ,but you've kind of confused me with "Consider an element radius r, width dr, length drθ" Is there another way to explain this part, I don't recall my Professor ever even mentioning the width being involved in anything, it was always the radius.
 
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Alright so the Force becomes F=Mg ,but you've kind of confused me with "Consider an element radius r, width dr, length drθ" Is there another way to explain this part, I don't recall my Professor ever even mentioning the width being involved in anything, it was always the radius.
I mean an element of the disk surface. Are you familiar with integration using polar coordinates?
 
  • #5
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I mean an element of the disk surface. Are you familiar with integration using polar coordinates?

I am not. Which I assume why I'm confused.

On another note, I've been trying to think of a way to figure this problem out, and I think for the first part I'm trying to go with using T=Iα however I'm not sure what α is. I have a hunch though, that because its the only force making the object come to a stop, I kind of remember there being a relationship between deceleration and friction force. Maybe something like α= -μkMg which is just having the deceleration equal the friction force? I mean it makes sense to me, but I don't know if its correct.
 
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haruspex
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I am not. Which I assume why I'm confused.

On another note, I've been trying to think of a way to figure this problem out, and I think for the first part I'm trying to go with using T=Iα however I'm not sure what α is. I have a hunch though, that because its the only force making the object come to a stop, I kind of remember there being a relationship between deceleration and friction force. Maybe something like α= -μkMg which is just having the deceleration equal the friction force? I mean it makes sense to me, but I don't know if its correct.
The obvious way to solve it is with integration. The methods you propose will not work.
Maybe there's another way, but I don't see it. I can step you through it if you wish.
 
  • #7
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The obvious way to solve it is with integration. The methods you propose will not work.
Maybe there's another way, but I don't see it. I can step you through it if you wish.

If you could, that would be greatly appreciated. Why wouldn't my idea work though? Does the relationship between deceleration and the friction just not work that way?
 
  • #8
haruspex
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If you could, that would be greatly appreciated. Why wouldn't my idea work though? Does the relationship between deceleration and the friction just not work that way?
##\alpha=\mu_kMg## is dimensionally wrong. On the left is an angular acceleration, on the right a linear acceleration. You do need to find the total torque.

Consider a narrow annulus of the disk at radius r, width dr. Of that, consider a short section subtending an angle ##d\theta## at the disk's centre. The length of that section is ##r.d\theta##. The area is ##r.d\theta dr##. Since the mass is uniformly distributed, its weight is ##g\rho r.d\theta dr##, where rho is the mass per unit area of the disk. The frictional force on the element is ##\mu_k g\rho r.d\theta dr##.
That frictional force is directed along the tangent to the circle radius r.
What is its torque about the centre of the circle?
What is the torque of the entire circular element radius r, width dr? (I.e. extending ##d\theta## to be 2 pi)
 
  • #9
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##\alpha=\mu_kMg## is dimensionally wrong. On the left is an angular acceleration, on the right a linear acceleration. You do need to find the total torque.

Consider a narrow annulus of the disk at radius r, width dr. Of that, consider a short section subtending an angle ##d\theta## at the disk's centre. The length of that section is ##r.d\theta##. The area is ##r.d\theta dr##. Since the mass is uniformly distributed, its weight is ##g\rho r.d\theta dr##, where rho is the mass per unit area of the disk. The frictional force on the element is ##\mu_k g\rho r.d\theta dr##.
That frictional force is directed along the tangent to the circle radius r.
What is its torque about the centre of the circle?
What is the torque of the entire circular element radius r, width dr? (I.e. extending ##d\theta## to be 2 pi)

I feel really bad because you're taking the time to help me, but this method is just completely new to me. I've checked with some class mates, and my class wasn't taught this, so it wouldn't be accepted. I'm so sorry, but are you sure there's no other way to go about this problem?
 
  • #10
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I feel really bad because you're taking the time to help me, but this method is just completely new to me. I've checked with some class mates, and my class wasn't taught this, so it wouldn't be accepted. I'm so sorry, but are you sure there's no other way to go about this problem?
Try this: consider a narrow triangle with its apex at the centre of the disk and its base at the circumference. You can treat the frictional force as being approximately parallel to the base everywhere in the triangle. The normal force is uniformly distributed over the triangle, so the frictional force is too. That means you can treat the friction as acting at a particular point within the triangle.... what point?

(There is another interpretation of the problem. Maybe the disk is vertical. But then you need to assume the axle is fixed so that it cannot roll, so I doubt that's it.)
 

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