And this problem were given a position vector as a function of time. Um that describes the motion of a particle of mass three kg. So the position vector is a is a again a time dependent vector. And they asked us first for the um for the torque acting. Um Well they ask us for the torque acting on the particle. Alright, sub, there's a couple ways we could do that. Um We can either say that. Now. One thing you have to be a little careful about is that you can't just say the torque. Um Let's see here, I'll do this in red because this is not correct. The torque is not equal to our cross em eh That is not correct because what is correct is the torque equals DDT of our cross M. B. The angle madam. And then we have to use the product rule. So that is um um we can pull the em out if that's not a function of time. And so then we get V Cross with V. Plus M. R. Cross with a, well what we see here then is that we actually get V cross with V is zero. Right? So it turns out that in fact the torque is this, so this is actually correct. Um But you can't just you got to remember when you that you have to remember that you're taking this whole derivative here. So a couple of things we could do we could either calculate the acceleration across that with our or calculate the angular momentum first and then um and then take the time we were devoted and that's what I did. Again it's either way you should get the same answer. But again it's well this is correct. You've just got to remember that T. R. Is a function of time too. So it is correct just because it happens to be that you get re crossed with me here and that term drops out. But again, just writing this down naively, um might get you in trouble. Um So let's see here, we have um our angle momentum, our cross with the So we worked through the cross product and that turns out that the stuff cancels out here, which is nice. So it turns out that the angle momentum is simply in the K direction and it's 2040 squared in the Z. Direction and then the units are kilogram meters squared per second. So that's nice because the cubic term dropped out. Well it's not really nice just we could do the problem. That wasn't the case. But we just get the we just get the square term. So That's because we have 48 t. cubed here and then we have 48 here and then there's ones up canceling out. So that makes this fairly easy, so we just to take to find the time from time derivative of the angle momentum. Um that's just 4080 in the Z. direction. And that has a unit of Newton's nuclear tens, meters. Um Obviously we've converted some stuff here to Newton's, so let's see here. They wanted the magnitude, so the magnitude of the angular momentum Is 24 he squared. And so that's obviously increasing in time. So where whatever this whatever forces causing there's some kind of force or torque acting on the particle. Well, if it's a particle there's some kind of force acting on it that's generating a torque that's causing the angular momentum to increase with T. Squared.