00:01
This is a dynamics rotational problem.
00:04
You have an object on a merit around a distance of 5 .4 meters from the point of rotation.
00:11
You've got to try and solve for the force that will keep it from sliding off.
00:15
So there's a couple things that work here.
00:16
You need to do the equation for friction.
00:18
It's not moving, so it's static friction.
00:20
So ff equals mu s times fn, or fn is the normal force.
00:27
Because it's sitting on a horizontal surface, the normal force is going to equal the force of gravity, which is going to be mg.
00:35
So i'm going to leave that in variable form for now.
00:38
That frictional force, this is the object as seen from the side view.
00:42
That frictional force is what is pulling it in towards the center of rotation.
00:46
So this is also going to be the centripetal force that is keeping the object on the ride, and that's calculated by mv squared over r.
00:58
They want to know the coefficient of static friction in us that will make that happen given a term rate of one revolution per six seconds and a radius of rotation of 5 .4 meters.
01:12
So you're going to need to know that the velocity v of the object rotating is 2 pi r over the period or 2 pi times 5 .4 over the period, which is 6 time for one revolution...