00:01
In this problem, we're going to talk about angular momentum and torque.
00:04
So first, remember that the angular momentum l associated with rotating particle is equal to the mass of the particle times the radius of rotation times the linear accelerate, i'm sorry, the linear speed v.
00:25
Also, remember that the torque is equal to the change in angular moment.
00:34
Divided by the change in time.
00:37
So it's delta l divided by delta t.
00:40
And finally what we're going to need to our exercise is that the work exerted can be calculated as the torque exerted over a particle times the variation in the angular position of that particle.
01:02
So now i can go into our problem.
01:04
We have a water wheel that has a radius of 3 meters.
01:10
And we have water that enters the water wheel with an initial speed of 7 meters per second and exits the water wheel with a final speed of 3 .8 meters per second.
01:27
And in question a, we have to consider that the flux of water through the water wheel is 85 kilograms per second.
01:36
And we have to calculate what rate, i'm sorry, what is the rate with which the water delivers angular momentum to the water wheel.
01:53
So the water loses angular momentum and this angular momentum ends up with the water wheel.
02:00
Okay, so let's calculate what is the loss in angular momentum of the water? notice that the initial angular momentum is the mass of the water.
02:11
This actually is the mass divided by time.
02:14
So this is the mass of the water times the radius of the water wheel times the initial velocity, while the final one is the mass times the radius times the final velocity.
02:26
So the change in angular momentum is mr times vf minus vi.
02:34
Okay, so this is 85 kilograms.
02:38
In one second, we're going to calculate the change in angular momentum per unit time.
02:45
So it's 85 kilograms per second times three meters times 7 minus 3 .8 that is 3 .2 meters per second...