00:01
For this problem on the topic of angular momentum, we have a string which is wound around a uniform disk, which has a radius r and mass m.
00:07
The disc is released from rest when the string is vertical and its top end is tied to a fixed bar as shown in the figure.
00:14
We want to show that the tension in the string is a third of the weight of the disc.
00:19
The acceleration of the center of mass has magnitude 2g over 3, and the center of mass has speed 4gh over 3 all to the power half.
00:28
We then want to verify this answer using the energy approach.
00:33
So if we take the sum of the forces on the disk, this is equal to the tension t minus its weight m g.
00:42
And by newton's second law, this is minus m a.
00:49
If we take the sum of the talks, we have this to be t times r.
00:56
And again, by newton's second law for angular motion, this is i times alpha.
01:01
Where i is the moment of inertia and alpha the angular acceleration.
01:05
And so for a disk, we can write this as a half m r squared times a over r.
01:16
Now, if we combine these two equations, we find that the tension t is equal to m into g minus a.
01:30
And a is equal to 2t over m.
01:39
And so from here we can see that the tension t is equal to mg over 3.
01:51
That's a third of the weight of the disk as required...