00:01
So we have a cylinder rolling on a cart that is being pulled with a force t.
00:07
So the first thing we're going to do is we're going to, i've listed all our givens here and converted the weights into masses.
00:15
So these are, this is mass a, this is mass b, this is the radius and feet, this is the pulling force, and we have a t of 1 .2 seconds.
00:27
So let's examine this rolling motion of this cylinder.
00:35
So we have a normal force being exerted.
00:41
We have a frictional force being exerted.
00:44
And we have a m .g that's being exerted directly downwards along the center.
00:51
And this is all around a point c.
00:54
So if we have moments about c, you'll notice that there are no, there is no impulse being imparted upon this cylinder because none of these forces create a moment.
01:19
So we have 0 is equal to i omega, or i'm just going to say 1 half m r squared omega minus m -a -v -a -r.
01:31
And this is the linear component of the momentum, and this is the rotation.
01:37
So that means that we have 1ā2 .559 times 0 .75 squared omega minus 0 .559 times 0 .75va.
01:58
So we get 0 .157 omega minus 0 .419 va.
02:27
So now we have that.
02:32
We are going to be, we're going to use this as one of these simultaneous equations that we will use to solve this problem.
02:46
So we're going to move this up here and continue.
02:53
So next we have, we're going to look at the cart, the cart pulley system as a whole.
03:06
And note that there is an impulse being created called pt, that the cart is also moving with some mbvb...