00:01
So we need to find the rate of the angular momentum of the wheel.
00:04
And then we have the torque on the wheel and then the power generated.
00:10
So we have the radius of the wheel 3 .0 meters.
00:13
And then we have v .1 equaling 7 .0 meters per second.
00:18
V sub 2 equaling 3 .8 meters per second.
00:23
We know that mass divided by t.
00:25
So the amount of water going through the wheel every second, 85 kilograms per second.
00:33
So for part a, we consider that the delta, the change in the angular momentum of the wheel will equal the change in the angular momentum of the water.
00:44
So this will be equal to delta l initial of the water.
00:51
And then this will be minus delta l final of the water.
00:57
And this can be equal to mv1r minus mv2.
01:03
Are given that omega equals vr so rather given that um l rather the angular momentum equals i omega so at this point we can say that um delta l of the wheel divided by t delta t would be equal to m delta t times r times v sub 1 minus v sub 2 so we can say that this is going to be equal to 85 kilograms per second.
01:43
So that'll be m divided by delta t and then times the radius of three and then times seven minus 3 .8...