Calculating Torque for Rotating Disk on Flat Surface

In summary, the conversation discusses finding a formula to calculate the torque needed to rotate a flat disk on a flat surface with a known coefficient of friction between the two materials. The suggested formula takes into consideration the friction between the surfaces, speed of rotation, radius or area of the contact surface, and the torque required to start the disk spinning. It involves using calculus to divide the disk into infinitesimal rings and integrating over the actual radius to get the total torque required. The final formula is \tau= {2\over 3}F\mu R, where \tau is the torque, F is the total force exerted against the glass, \mu is the coefficient of friction, and R is the radius of the disk.
  • #1
Mightymuff
4
0
Ive been trying to find formulas that help describe the torque needed to rotate a flat disk on a flat surface when the coefficient of friction between the 2 materials is known.

The first surface is a glass window (so a vertical plane) the second surface is a rotating disk of wet sponge "cleaing it" (also a vertical plane).

So I am looking for a formula that takes into consideration friction between the surfaces, speed of rotation, radius (or area) of the contact surface and torque required to start the disk spinning.

Any suggestions?
 
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  • #2
If these materials are "normal", then speed of rotation will probably not affect the torque needed to keep it rotating. You're going to need some calculus to solve this problem.

You basically have a disc that can be divided into infinitesimal rings of area dA proportional to a mass dm. Each ring contributes some frictional force dF applied at a distance r. Integrating over the actual radius of the disc, you will get the total torque required.
 
  • #3
mezarashi said:
If these materials are "normal", then speed of rotation will probably not affect the torque needed to keep it rotating. You're going to need some calculus to solve this problem.

You basically have a disc that can be divided into infinitesimal rings of area dA proportional to a mass dm. Each ring contributes some frictional force dF applied at a distance r. Integrating over the actual radius of the disc, you will get the total torque required.

Thanks for the response but my maths is extremely rusty. I understand the general concept of your response but what would the exact formula look like when complete so that i could use it?
 
  • #4
You can make this as easy or complicated as you like. I personally start as easy as possible and then introduce the complications based on how accurate I need to be.

In this case, why not estimate your frictional force and use the relation

[tex]\Sigma T = I \alpha[/tex] ?

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
 
  • #5
FredGarvin said:
You can make this as easy or complicated as you like. I personally start as easy as possible and then introduce the complications based on how accurate I need to be.

In this case, why not estimate your frictional force and use the relation

[tex]\Sigma T = I \alpha[/tex] ?

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Is that not moment of inertia?

The idea is to have an equation that works out the torque required to rotated the vertical disk. It must include the coeficient of friction between the surfaces of the disk/window as one of the variables. I assume the torque needed to overcome the friction between the surfaces would be more than the moment of inertia.

Maybe the total torque would be moment of interia + the frictional force it has to overcome from the 2 materials rubbing?
 
  • #6
The final formula is:
[tex]\tau= {2\over 3}F\mu R [/tex]

[tex] \tau [/tex] is the torque.
[tex] F [/tex] total force exerted against the glass.
[tex]\mu [/tex] the coefficient of friction.
[tex]R [/tex] the radius of the disk.
As said, it independent of the speed of rotation.
 
  • #7
lpfr said:
The final formula is:
[tex]\tau= {2\over 3}F\mu R [/tex]

[tex] \tau [/tex] is the torque.
[tex] F [/tex] total force exerted against the glass.
[tex]\mu [/tex] the coefficient of friction.
[tex]R [/tex] the radius of the disk.
As said, it independent of the speed of rotation.

Thank you very much that's perfect.

Could you please take a couple of minutes to explain how you got that formula though please.
 
  • #8
I gave you the solution because I thought that you are not a physics student but that you need this formula for your job. If this is the case, you do not need the derivation.
If you are a physics student then do as mezarashi told you (this is, of course, what I did).
 
  • #9
Mightymuff said:
Thank you very much that's perfect.

Could you please take a couple of minutes to explain how you got that formula though please.

Sorry for the necro.

Was working on a similar problem today and devised this integral. After forgetting how to do integrals, I thought my answer was wrong, and then I came here. Seeing the final answer reminded me how to do calculus.

Here is a picture trying to show some of the math behind it.

mLbBb.jpg


Edit: and sorry for the same step being in there twice.
 
Last edited:

1. What is the equation for the rotational speed of a disk?

The rotational speed of a disk is given by the equation ω = v/r, where ω is the angular velocity in radians per second, v is the tangential velocity in meters per second, and r is the radius of the disk in meters.

2. How does the equation for rotational speed change if the disk rotates at a constant acceleration?

If the disk rotates at a constant acceleration, the equation for rotational speed becomes ω = ω0 + αt, where ω0 is the initial angular velocity and α is the angular acceleration in radians per second squared.

3. Can the equation for rotational speed be used for non-uniformly rotating disks?

Yes, the equation ω = v/r can be used for non-uniformly rotating disks as long as the tangential velocity v is known at each point on the disk's surface.

4. Is there an equation that relates the rotational speed of a disk to its moment of inertia?

Yes, the equation for rotational kinetic energy, K = 1/2Iω2, can be rearranged to solve for ω and gives ω = √(2K/I), where K is the rotational kinetic energy in joules and I is the moment of inertia in kilograms per square meter.

5. Can the equation for rotational speed be used for disks with a non-circular shape?

Yes, the equation ω = v/r can be used for disks with a non-circular shape as long as the tangential velocity is measured at the point on the disk's surface that is at a distance r from the center of rotation.

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