Time taken to stop

Question

A disc of radius R is rotating on a rough surface about its natural axis with an angular frequency w. Assuming the coefficient of friction of the floor to be u, how to find the time taken by it to come to rest?

Solution

If we can find the torque due to friction our work is done.

The area density of disc !!=M/(piR^2)!!

To find the torque due to friction. Consider a ring of width !!dx!! and radius !!x!!.

The tangential force acting on the rim will be mass of ring multiplied by coefficient of friction and gravity so we get !!2pix xx dx xx M/(piR^2)xx mu xx g!!

So the torque on the ring due to the friction is !!x xx 2 (xM)/R^2 muxxgdx!!

So the total torque !!=int_0^R (2muMg)/R^2 x^2 dx!!

!!=(2muMg)/R^2 R^3/3!!

!!=2/3 muMgR!!

Now the time taken would be:

!!t=omega/alpha=omega/(tou/I)!!

!!=(omega(MR^2)/2)/(2/3 muMgR)!!

!!=3/4 (omegaR)/(mug)!!

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