Slipping and rotation

I know this is a bit late, but if you still haven't figured it out, here goes:

Firstly, let the mass per unit area of the disk be $\sigma =\frac{m}{\pi R^2}$.

Consider a small element of area $dA$ at a distance $x$ from the center of the disk. The mass of this element is $dm=\sigma dA$. The frictional force on this element is $dF=\mu (dm)g=\mu \sigma g\times dA$

The torque on this small element is $d\tau =x\times dF=\mu \sigma gx\times dA$. (note that by $\times$ I just mean normal multiplication, not the cross product or anything).

Now let us consider a thin circular ring with thickness $dx$ at a distance $x$ from the center (I mean to say a ring of radius $x$ whose center is the center of the disk). The net torque $\tau$ on this ring would be the integral of all the small torques on the small elements of area $dA$ lying on this thin ring.

$$\tau =\int d\tau =\mu \sigma gx\int dA=\mu \sigma g\times (2\pi x^{2}dx)$$

Note that in this integral $x$ is a constant, since $x$ is the distance from the center for all the small elements on the circular ring we have taken under consideration.

As $dx\to 0$, we have infinite such rings on the disk. So now we need to integrate over all these rings. The equation above matches the second equation of your post.

I think you should be able to take it from here. The important thing is to keep in mind what we are doing: first we find the net torque on a thin ring situated at a distance $x$ from the center which I've done above, and then we integrate over all such rings.

posted over 3 years ago

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TripleFault‭

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