consider a rotating cylinder with a man in it standing against the wall. The floor is removed.I have to find the minimum velocity for which the man will continue to stay that way(not fall). My question is why do we have to take the max/limiting value of friction?

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- anonymous

- katieb

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- anonymous

|dw:1341053885107:dw| v is the velocity r is the radius now and \[\mu \] is the cofficient of friction now \[v ^{2}/r\] will be the acc and the force willl be \[\mu m v ^{2}/r\] here mv^2/r is due to the cylinder so has to be adjusted to the max so that it counteracts mg so friction will be max

- anonymous

* so \[\mu \] has to...

- anonymous

why cant less than max friction counteract mg?

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## More answers

- anonymous

we need to take max limiting value of friction here because you need to find the " minimum velocity" .

- anonymous

a force needed to stick that man to the wall is constant if you choose one of the variable to be the max then other has to be min and vice versa. you want min velocity you must ake max frictional constant.

- anonymous

i still dont understand how the friction has limiting value

- anonymous

can u xplain mathematically?

- anonymous

mathematically \[F =muv ^{2}/r\]
so \[\mu \alpha 1/v\] or
mu is inversly proportional to velocity thus if you want minimum velocity then you must consider max friction constant or max friction offered.

- anonymous

but how can we write
F=mu mv^2/r in the first place
??

- anonymous

why the man isn't inside?

- anonymous

?

- anonymous

i mean that friction can be LESS THAN or equal to that value right? why do we take it equal?

- anonymous

|dw:1341414120875:dw|

- anonymous

friction can be LESS THAN or equal to that value right? why do we take it to be equal?

- anonymous

friction force never will be higher than mg,to be in equilibrium,it must me equal.

- Vincent-Lyon.Fr

This is a case (inequalities) where you need to 'feel' which way things go.
Try to understand imagining that you start with a very HIGH angular speed :
Normal force will be very big and, as friction adapts to other constraints, it will just be equal to weight, which will be much less than its maximum value of µN. No problem, the body will stay in equilibrium.
As you decrease the speed, friction will remain same (opposing weight), but as normal force will decrease too, there will be a time when friction exactly equals µN. (boly still at equilibrium)
This will be the limiting (minimum) value for omega. If you decrease omega further, then, in no way will friction remain inferior to µN, and the body will start slipping.

- ujjwal

Here \[mg=\mu \frac{mv^2}{r}\]or, \[v=\sqrt\frac{rg}{\mu}\]In above relation, v and \(\mu\) are inversely proportional to each other.
So, clearly, for minimum velocity, \(\mu\) has to be maximum.
Or, you can understand it theoretically in this way:
when v is less, centrifugal force is less and hence the normal reaction to wall is less. So, you need a greater value of \(\mu\) to provide necessary friction so as to balance the weight (which is always constant).
Hence for minimum velocity, \(\mu\) has to be maximum.

- anonymous

its actually too simple, because the rotation is throwing him outwards on the walls of the cylinder but, nothing is stopping him from falling down under the effect of gravity (when the base is removed). so we should have a limiting value of friction that will save him from the gravity......

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