Understanding Friction Between Two Disks: A Problem in Rotational Dynamics

[JD_PM]

Homework Statement

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?

 

[oz93666]

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.

 

[haruspex]

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.

 

[haruspex]

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.

 

[JD_PM]

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

 

[haruspex]

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.

 

[JD_PM]

I do not understand where this equation comes from. There are no linear accelerations, only angular accelerations.

Yeah, my bad thanks.