Torque friction on a rotating disk

[kudoushinichi88]

A uniform horizontal disk of mass M and radius R is rotating about its vertical axis with an angular velocity $\omega$. When it is placed on a horizontal surface, the coefficient of kinetic friction between the disk and the surface is $\mu_k$.Find

a)the torque $\tau$ exerted by the force of friction on a circular element of radius r with width dr.

Let the element has a mass of dm.

$$\frac{dm}{M}=\frac{\pi(r+dr)^2-\pi r^2}{\piR^2}$$

$$dm=\frac{2M}{R^2}rdr$$

The torque is provided by the frictional force, therefore

$$d\tau=f_{friction}r=\mudm\ gr$$

$$d\tau=\frac{2M\mu g}{R^2}r^2dr$$

b)the total torque exerted by friction on the disk

Integrating,

$$\tau=\frac{2M\mu g}{R^2}\int_{0}^{R}r^2dr$$

gives us

$$\tau=\frac{2}{3}M\mu gR$$
c)the total time required to bring the disk to a halt.

$$\tau=I\alpha$$

Since disk is decelerating,

$$\alpha=-\frac{\tau}{I}=-\frac{4\mu g}{3R}$$

$$\omega=\omega_0+\alpha t$$

$$t=\frac{3R\omega}{4\mu g}$$

Can someone check my work? I don't have solutions to this question...

 

[Delphi51]

Every line looks great to me!

 

[kudoushinichi88]

Thanks! I'm not sure of my work because I'm not sure if I understand how torques work on surfaces right...

 

[Delphi51]

No problem with the physics.

 

[Delzac]

I have a question, how was the 1/2 in

$$\frac{dm}{M}=\frac{\pi(r+dr)^2-\pi r^2}{\piR^2}$$

obtained?

And by the way, this is the dm for a circular ring? If i am not wrong?

 

[Delzac]

And how did you simplify to the step below of the first?

 

[kudoushinichi88]

Oh, that is a typo.. the 2 should have been the total area of the circle, $$\pi R^2$$.

To simplify the equation, you must realize that dr^2 is so small that it could effectively be taken aa 0.

 

[Delzac]

Yeah got it.

 

[Tyvan07]

I'm having trouble adapting this calculation to solve my problem. Instead of the mass of the disc acting as friction source, I have an axial force applied to a hollow shaft and the mass can be neglected. Here's an illustration...coffee cup upside down on desk (assumed wall thickness) and pressed down with some force, how much torque is required to 'break' static friction?

Any takers? I'll try to tackle it myself and post what I come up with.