In fig 2 a small block of mass m is placed on the inner wall...

In Fig. 2, a small block of mass m is placed on the inner wall of a hollow thin cylinder of radius R = 20.0 cm, which rotates from rest about a vertical axis with acceleration of 2.20 rad/s2 . After 5.50 s, the block begins to stay on the wall at a constant height. (a) Find the coefficient of static friction between the block and the wall of the cylinder. (b) How many revolutions has the cylinder turned after this interval of 5.50 s?

Transcript

00:01 So let's say we have a cylinder, a hollow cylinder that can rotate with some angular velocity omega.

00:09 And a small block rests against the side of the cylinder once it reaches a certain speed.

00:16 But the cylinder has a radius of 0 .2 meters.

00:21 So this is r.

00:22 And we're told that it rotates from rest to or with an acceleration, of 2 .2 radians per second squared for 5 .5 seconds after which you know that's when the block becomes stationary and so we want to find first what is the coefficient of friction between the block and the cylinder and then how many revolutions the cylinder has turned okay so what we're going to have is if we think about a free body diagram of the block this is our block we're going to have gravity of course going straight down and we're also going to have the normal force from the wall going this way, which is really our centripetal force.

01:12 So depending on how you want to label that, maybe we'll write it as a normal force.

01:16 And then, of course, we're going to have our frictional force.

01:18 I'll write us with the lowercase f.

01:22 So we know the frictional force is going to be the coefficient of friction, coefficient of static friction, times the normal force.

01:29 And in this case, the normal force is equal to the centripetal force.

01:32 Force.

01:33 So we can write this as mu s times m v squared over r.

01:40 Or alternatively, we could write it as m omega squared times r, or omega is the angular velocity at this point.

01:48 And at equilibrium, this is equal to the weight...