00:01
So this question deals with an at -woods machine with the additional factors that we have to consider this time of the mass of the pulley, meaning the pulley will have some rotational inertia.
00:15
And there is some friction in the pulley that we have to consider as well.
00:21
So as suggested in part a, here are the free body diagram showing all the forces acting on the three different objects.
00:30
On mass 1, there is a force of gravity pulling down of m1 times g and a tension force pulling up of t1.
00:42
For object or mass 2, there is a force of gravity pulling down of m2 times g and a tension pulling up of t2.
00:52
And then on the pulley itself, there are three forces that are under consideration that we need to consider because they might exert torques.
01:01
We have tension one pulling down on the one side as i've drawn it.
01:08
And then tension two on the other side as i've drawn it.
01:11
And then there's a force of friction which is going to exert a frictional torque that will need to consider in our torque equation.
01:18
So let us write equations from each of those drawings.
01:26
For the mass one, we can say, let up be the positive direction for mass one.
01:32
So therefore, t1 minus m1g is equal to m1 times a, where a is the acceleration of mass 1.
01:47
We can also then write for mass 2.
01:51
Now, for mass 2, in order to keep using the same acceleration for both objects, i'm going to say that down is the positive direction for mass 2.
02:02
That allows me to keep a and not have to change a sign when i put it in the equation for 2.
02:11
So for mass 2, the equation is going to be, it's going to be m2a, m2g, i should say, minus t2 is equal to m2a.
02:29
And then for the pulley, we have three forces that will exert torques on it.
02:41
And those forces are going to, those torques are going to be then.
02:46
We have t2 times the radius of the cylinder, because the rope is apparently wrapped on the outside of the cylinder, minus t1 times radius of the cylinder.
03:00
That's the torque exerted by the tension.
03:03
In rope 1.
03:06
And then, because it goes against the motions, we're going to say minus.
03:11
And this is the frictional torque that was given to us.
03:16
And by newton's second law for rotation, this should equal the rotational inertia of the pulley at times its angular acceleration.
03:25
Now, we can simplify this because we have several unknowns that we can combine some equations together.
03:31
All we can do is we can add the equations for t1 and t2 together.
03:39
And then we can substitute that equation.
03:42
It'll give us an equation that has t1, or t2 minus t1.
03:48
We can substitute that in the equation for the rotational motion.
03:53
So in other words, t1, according to our first equation up there, is equal to m1a plus m1a, g whereas t2 according to the second equation we have up there is equal to m2g minus m2a and so what we're going to do is we're going to put it in the equation in the equation for the for the torques well for the torque equation we could divide out by r, that would help us a little bit.
04:51
So this becomes t2 minus t1 minus the frictional torque divided by the radius is equal to i times alpha divided by the radius.
05:17
And so we have here, now t2 is equal to m.
05:24
M2g minus m2a, so we can write that in here.
05:27
This gives us m2g minus m2a minus the factors that are in m1, so this is going to be minus m1a and then minus the frictional torque divided by r.
05:57
And let's work on this i term here.
06:02
I is assessing the problem is equal one half, big m, r squared...