00:01
All right, so we need to solve this using energy.
00:04
And so what we're going to need is a change in energy for the hanging mass.
00:13
So the hanging mass here will experience a loss of potential energy that will be equal to its mass, so 2m, times g, times the distance that it drops.
00:28
So it's going to lose that amount of potential energy.
00:33
Then we're going to set equal to the kinetic energy, both rotational and linear, for all three things.
00:41
So starting with the 2m hanging mass, we're going to have a kinetic energy when it reaches d of one half times its mass, so 2m, times its speed squared.
00:57
We'll also have a linear kinetic energy from the roller.
01:01
Cylinder and so that will be one half times its mass which is just regular m times that same speed squared.
01:12
Next we'll have a rotational kinetic energy of the rolling cylinder which will be one half times i times omega squared where i is the rotational inertia and we'll have a rotational kinetic energy of our pulley which would be one half.
01:30
Well, let's call these one and then we'll have one half i2 omega 2 squared for the pulley now i 1 is a cylinder so it's one half m r squared but this r is 2r so that we're going to put in as 2r quantity squared and i2 is also a cylinder so that's one half mr squared and this one has a radius of r, so it'll be one -half and r squared.
02:08
Omega 1, the rotational speed of that cylinder, assuming it rolls without slipping, will be v over r, which since we're squaring it, and that is 2r for the large cylinder, and since we're squaring that, this will be v squared and 2r squared, and omega 2 .2 will be the same thing, v over r, except this one has a radius of r.
02:38
So when we plug these in, we'll get one -half times one -half m2 -r quantity squared times v squared over two -r quantity squared.
02:47
So the two -rs are going to cancel.
02:49
And we have one -half times one -half, so that becomes one -fourth...