Friction Torque on a Rotating Disc

[mlab]

thread404-264188

On the thread listed above I was looking at it and had a question on how the final formula (T=mu*F*R*2/3) was formulated.

when i take the double integral of r*dr*dthetha with the bounds of 0 to R and 0 to pi I get pi*R^2/2.

I take P = F/A and A = pi*R^2/4

so substituting I get T=2*mu*F*R

Can someone help me on this.

Thanks,
Matt

 

[3DDave]

Part of the derivation is given as rdA where dA = rdr, so the equation is r^2dr -> r^3/3.

 

[mlab]

sorry A = pi*r^2, so still using r^3/3 and substituting back in would still only give you 1/3*F*R*mu.