Hi, I'm going to use the rotational analog to Newton's second law to solve the following problem. So we have a vertical post which is attached to a frictionless pivot point and this person comes along and pulls on this rope which is attached to the top of it. The exert a force or attention force of 400 newtons. The post is perfectly vertical, has a length of 3 meters and the mass of the post is 80 kilograms. And ultimately what I'm going to calculate is the angular acceleration. That's what this is going to result in. There are also a couple other skills that I'm going to practice here. One is calculating torque. The other is calculating momentum inertia. Once I have those two then I'll use the rotational analog to Newton's second law to calculate the angular acceleration which is the ultimate goal here. So torque from this tension force will be just like every other torque we can use the definition. So distance from the axis of rotation times the force times the sine of the angle between the r vector and the force vector. This is a tricky part. Okay, so r is very easy. It's the distance from the axis of rotation to where the force is exerted in this case. It's the full length of the pole. So I can just throw in three there. The force in this case is the tension force of 400 newtons. So I just throw that number in as well and then I need to figure out this angle. Now I intentionally do this angle to be not the correct angle to put into this formula. We need the angle between the r vector and the force vector. So one thing I do in these types of problems is I redraw the r vector and the force vector and then I determine this angle. So the r vector goes from the axis of rotation to where the force is exerted. So the r vector will point up. There's my r vector. And then my force vector points down 45 sorry 55 degrees below the positive x axis. So the angle that I need to put in there is this one right here. So it's not 55. In fact it's not minus 55. I'm going from r to f clockwise. So that's going to be a negative number and it's 90 plus 55. So negative 145 degrees. I throw that into my calculator and it tells me the torque is minus 688 Newton meters. OK, so that's the torque. There are other forces actually on this but they don't exert any torques. So there is a force at the pivot point, the normal force pointing up. But because it's exerted at the pivot point r is zero so it's irrelevant. There's also a weight force down which is opposite the r vector and we get a sign of 180 degrees which is zero. So neither the normal force nor the weight force exert torque. So this is the total torque. I can just say that summation of the torques is equal to minus 688 plus zero plus zero. OK, so I have that. I'm halfway to being able to use the rotational analog to Newton's second law. The other half of it is to calculate the moment of inertia for this post. And so this is typically something that you'll look up in a table depending on the geometry based on the pivot point and the shape of the object. And for this particular shape, a beam or post or whatever with an axis of rotation at one end has a moment of inertia of one third times the mass times the length squared. And so just chuck numbers into that formula and you get the moment of inertia. So be one third times the mass of 80 kilograms times the length of three and you might be able to do this one in your head. It is 240 kilogram meters squared. OK, so that's the moment of inertia for this particular object. And now we can use the rotational analog to Newton's second law. So the rotational analog to Newton's second law looks like this. Summation of the torques equals the moment of inertia times the angular acceleration. And then I can solve for angular acceleration total torque divided by moment of inertia. I've got both of those numbers. And so the torque of negative 688 divided by the moment of inertia of 240. And I throw that into my calculator and it spits out an answer of negative 2.87 and SI units. Since I put SI in, I'll get SI out to negative 2.87, 7 radians per second squared. What does the negative sign mean? It just means that I'm inducing a clockwise angular acceleration. One thing to note about this is that as soon as this begins to spin, all of a sudden the weight is going to actually be exerting a torque depending on how this person moves and holds the rope, the angles of that might change. So a lot of things could change. So I wouldn't assume that this angular acceleration is constant. I therefore wouldn't try to use this number to predict without more data how quickly this thing is going to slam into the ground using constant angular acceleration rotational kinematics equations. But I certainly can assert that under these physical conditions, its instantaneous angular acceleration would be this number. So again, definition of torque, an order calculate torque, formula for this particular geometry to calculate momentum of inertia, followed by the rotation line log to Newton's second law in order to calculate the angular acceleration. OK, so thanks for watching.