Understanding Friction Between Two Disks: A Problem in Rotational Dynamics

In summary: The equation is from Newton's Second Law of Motion. It states that the force on an object is the product of its mass and its acceleration.
  • #1
JD_PM
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Homework Statement



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This problem was originally posted on Physics Problems Q&A: http://physics.qandaexchange.com/?qa=616/friction-between-two-disks

Homework Equations



Second Newton's law for rotation:

$$\tau = I \alpha = RF$$

The Attempt at a Solution



I tried to solve this problem as follows:

Using Second Newton's law for rotation on disk 2R:

$$2Rf = I \alpha = \frac{1}{2} M (2R)^2 \frac{a}{2R} $$

Plugging the numbers into the equation we get:

$$f = 2N$$

OK at this point, now let's do the same on disk R:

$$Rf = I \alpha = \frac{1}{2} M (R)^2 \frac{a}{R} $$

Plugging the numbers into the equation we get:

$$f = 1N$$

So I get different values for ##f## while I have to get the same value! What am I missing on my calculations on disk A?
 

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  • #2
I don't see why the moment of inertia of A comes into it , or it's radius , or mass...

The force on B is applied at the point of contact , and it's the inertia of B that resists this force.
 
  • #3
JD_PM said:
OK at this point, now let's do the same on disk R:
There are two torques acting on that disc. Your equation only shows one.
But your problem with this approach is you cannot evaluate the other torque without first finding the frictional force the way you did.
 
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  • #4
oz93666 said:
I don't see why the moment of inertia of A comes into it , or it's radius , or mass...

The force on B is applied at the point of contact , and it's the inertia of B that resists this force.
JD_PM has already answered correctly the question as posed, but is wondering why a different approach seems to give a different answer.
 
  • #5
haruspex said:
There are two torques acting on that disc. Your equation only shows one.
But your problem with this approach is you cannot evaluate the other torque without first finding the frictional force the way you did.

Yeah I see what you mean:

$$FR - fR = I \alpha = \frac{1}{2} M (R)^2 \frac{a}{R}$$

$$F - f = 1N$$

So F has to be 3N. (I used the value I got on disk 2R analysis)

But there is a problem: we know the acceleration and the mass of disk R; we can get F (using disk R):

$$F = M a = 1N$$

Mmm so there is something wrong here... F values are different using different approaches
 
  • #6
JD_PM said:
we know the acceleration and the mass of disk R; we can get F (using disk R):

F=Ma=1N​
I do not understand where this equation comes from. There are no linear accelerations, only angular accelerations.
 
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  • #7
haruspex said:
I do not understand where this equation comes from. There are no linear accelerations, only angular accelerations.

Yeah, my bad thanks.
 

1. What causes friction between two disks?

The force of friction between two disks is caused by the microscopic irregularities on the surfaces of the disks. As the two disks come into contact, these irregularities interlock and create resistance against movement, resulting in the force of friction.

2. How is the force of friction between two disks calculated?

The force of friction between two disks can be calculated using the formula F = μN, where F is the force of friction, μ is the coefficient of friction, and N is the normal force between the two disks. The coefficient of friction is a measure of the two surfaces' roughness and the normal force is the force exerted by one disk on the other perpendicularly to their surfaces.

3. Can the force of friction between two disks be reduced?

Yes, the force of friction between two disks can be reduced by using a lubricant between the two surfaces. Lubricants, such as oil or grease, create a barrier between the two surfaces, reducing the contact between them and thus reducing the force of friction.

4. How does the speed of the disks affect the force of friction?

The speed of the disks has a direct effect on the force of friction. As the speed increases, the force of friction also increases, making it more difficult to move the disks against each other. This is because the faster the disks move, the more contact points they have, resulting in a higher force of friction.

5. What are some real-world examples of friction between two disks?

Friction between two disks is a common occurrence in everyday life. Some examples include the brake pads and rotors in a car, the gears in a bicycle, and the record player needle on a vinyl record. Understanding the force of friction between two disks is important in engineering and design to ensure smooth and efficient movement between two surfaces.

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