00:01
Here we're going to do an example of conservation of energy in which one of the objects is going to gain some rotational kinetic energy.
00:15
So a couple of reminders is that if there's no friction or other ways to lose energy, your total mechanical energy is a sum of kinetic plus potential energies.
00:29
And if you have multiple objects in the system, you have to add in kinetic and potential of each object.
00:38
So here we have two objects.
00:42
The little mass, 15 kilograms that is hanging initially off the drum, has a kinetic energy equal to 1 .5m v squared.
00:54
And it also has potential energy in the gravitational field of m -g -y.
01:04
Meanwhile, the drum, so what to call this, little m, the drum, we'll just use subscript d for that, has rotational kinetic energy, and that is equal to one -half imega squared if we wanted to worry about what that is.
01:24
However, because of the way it is connected to the small mass, there is a common speed.
01:36
So omega is equal to the same speed of the little mass divided by r.
01:45
And we're never given the radius of the drum, so hopefully we're not going to need it.
01:51
So the idea is we're starting on earth.
01:55
And we're going to use earth to calibrate the system.
02:01
And what we know on earth is that when the little mass m falls 5 meters from rest, the drum winds up with kinetic energy of rotation equal to 250 joules.
02:36
Okay, so the idea is that we have, the energy of the system initial before everything drops is equal to zero.
02:51
And then finally, we've got the minus mg.
02:56
We'll pretend it starts at y equals zero.
02:59
That's always a good place to start.
03:05
So minus mg, and we'll call that for the earth times delta y.
03:11
Absolute value is equal to one half.
03:19
M v squared plus 250 joules.
03:31
Okay, and we would like to know from mars, if we were to drop it, how far would it have to drop in order to come up with the same 250 joules? and this is where it may be helpful to kind of write out the other side, as if it were a moment of inertia.
03:55
Let me see.
03:57
We would have plus i over r squared times v squared would be equal to minus m g e times delta y absolute value and what we can tell is that if we take this system to mars see so on mars what we have is, yeah, let's put a subscript on the delta y as well.
04:55
We have minus mg mars times delta y mars is equal to one half m plus i over r squared times v squared.
05:14
Okay, so the moment of inertia, the radius, the little mass, none of that will change when you go to mars.
05:23
So what we know is that if we take a ratio of the earth stuff to the martian stuff, basically we have to get one.
05:39
So in other words, m, g .e, delta y, earth over mgm.
05:50
Let's use blue there.
05:54
Delta y mars has got to equal one.
06:01
Nothing in that ratio is going to change for the drum.
06:07
And so to get 250 joules of energy into the drum, we basically have to have the same speed on mars as we do on earth...