Confusion about friction in rotating ring

[palaphys]

Homework Statement

A ring of mass m and radius r is rotating on a horizontal rough table with coefficient of friction. how long does the ring take to stop rotating? What is the frictional force from the ground?

Relevant Equations

Torque=r×F

So to begin with, I did this:

$$


\begin{align*}
dm &= \frac{M}{2\pi} d\theta \\
dN &= dm \cdot g = \frac{M}{2\pi} g \, d\theta \\
df &= \mu_k \, dN = \mu_k \frac{M}{2\pi} g \, d\theta \\
f &= \int df = \int_0^{2\pi} \mu_k \frac{M}{2\pi} g \, d\theta \\
f &= \mu_k \frac{M}{2\pi} g \int_0^{2\pi} d\theta \\
f &= \mu_k \frac{M}{2\pi} g \cdot 2\pi \\
f &= \mu_k M g
\end{align*}




$$

However, I had a thought- what if I could simply super impose all the frictional force vectors on every small element? Wouldnt that add up to 0?As there are vectors of equal magnitude in every direction.. Then why do I end up with some other answer?

 

[TSny]

However, I had a thought- what if I could simply super impose all the frictional force vectors on every small element? Wouldnt that add up to 0?As there are vectors of equal magnitude in every direction..

Yes. That's right. The net friction force is zero.

Then why do I end up with some other answer?

Your integration is adding up the magnitudes of the infinitesimal friction force elements. But, forces add as vectors, as you noted. So, your integration does not take into account the different directions of the infinitesimal forces.

This problem deals with rotation. What is the rotational analog of force?

 

[palaphys]

Torque

 

[palaphys]

So you say that due to the fact that vectorially the force is 0, the center of mass of the ring does not undergo any acceleration?

 

[TSny]

So you say that due to the fact that vectorially the force is 0, the center of mass of the ring does not undergo any acceleration?

Yes. I'm assuming the ring is lying "flat" on the table. The axis of the ring is vertical. Is this right?

 

[palaphys]

Yes

 

[TSny]

Torque

Yes. Can you see how to use torque in this problem?

 

[palaphys]

Yeah got it just gotta use tau is moment of inertia times alpha and then simple angular kinematics. Thanks

 

[TSny]

Yeah got it just gotta use tau is moment of inertia times alpha and then simple angular kinematics. Thanks

OK. Can you summarize how you found tau?

 

[Steve4Physics]

Can I briefly add to what @TSny has said?

Writing $f =\int {d\!f}$ is wrong and leads to an incorrect answer,

It should be $\vec f = \int \vec {d\!f}$. When we integrate a vector, the different directions of the elements must be taken into account.

 

[palaphys]

OK. Can you summarize how you found tau?

I used tau=I(alpha) =r×F

 

[palaphys]

Can I briefly add to what @TSny has said?

Writing $f =\int {d\!f}$ is wrong and leads to an incorrect answer,

It should be $\vec f = \int \vec {d\!f}$. When we integrate a vector, the different directions of the elements must be taken into account

If I did that would I get 0?

 

[kuruman]

If I did that would I get 0?

Yes, you would. Note that the torque on all elements $dm$ points in the same direction. Thus, if you add torque infinitesimals instead of force infinitesimals, you will have a non-zero answer.