00:02
Alright so for this question we're given the length of the bar.
00:04
I'm going to call that 2 meters.
00:07
The weight is 90 newtons.
00:11
Mass is given as 3 kilogram.
00:15
Alright so the initial velocity is 10 meters per second and the final speed is 6 meters per second.
00:24
So therefore because the separation between the rotational axis and the path of the ball is d, i'm going to have to call that 1 .5 meters.
00:37
Now the initial angular momentum about the pivot therefore, i'm going to have to call that ib1 should be equal to mu d.
00:49
Which means we're taking the mass 3 kilogram.
00:52
We're multiplying that by initial velocity of 10 meters per second and we're multiplying that by the distance 1 .5 meters.
01:00
So that we have 45 kilogram per square meter per second.
01:10
So therefore the final angular momentum of the ball, the pivot, will be equal to mvd.
01:16
So we're looking at 3 kilogram.
01:21
We're multiplying that by this 6.
01:24
We're also multiplying that by 1 .5.
01:27
And we have minus 27 kilogram square meter per second.
01:35
So therefore if we let angular velocity be that of the bar, therefore the moment of inertia of the bar in line with the pivot, okay, should be equal to m l squared over 3.
01:55
So you're looking at w l squared over 3.
02:00
So this is 90 multiplied by length squared and divided by 3.
02:07
And we're looking at g here.
02:09
So we're looking at 9 .8 meter per second squared...