Circular and Satellite Motion Legacy Problem #13 Guided Solution

My attorneys would advise me from the beginning to tell you, boys and girls, don't do this at home. But in 2002, professional skateboarder Bob Burnquist found himself a very large sewer pipe with a radius of about 1.8 meters. With his skateboard, he went inside the sewer pipe and began to rock back and forth along the walls of the sewer pipe. With one big burst of energy, he went around the full loop around that inside of that sewer pipe. Now, what we're asked here is what minimum speed would he need to have at the very top of the loop in order to make it through safely? Because you know that if he's upside down at the top of the loop and he's not going fast enough, that he's going to fall right on his head. So my solution to this problem, like many circular motion problems, will center around the use of a free body diagram. I draw a picture of a circle, or at least a part of a circle, the top part of a circle, and I put a dot at the so-called 12 o'clock position. At that 12 o'clock position, I think about the forces acting upon my object, the skateboarder. So I think there's always gravity force, and it is down at the very top of the loop. So F-grab is down, I draw an arrow down, and then I think he's upside down, so maybe the wall is pushing him down. So I could put F-normal down as well, an arrow. But here we're asked, what's the minimum speed which would be required at the top of the loop to make it through safely? And so what they're asking me is, what speed would this normal force be reduced to zero? For any speed less than this minimal speed, we wouldn't have circular motion, but instead would have projectile motion. So we're looking for the speed that gives us, as an F-net, simply F-grab. Now I won't do a lot of math in this problem, in fact, if you look at it, there's only one number to use, and that's the radius of the circle. So follow along as we talk through the algebra derivation of our equation. We begin with the idea that F-net equals F-grab. So if F-net equals F-grab, we could say that m a, that's how we write F-net, is equal to m g, that's how we write F-grab. So m's are on both sides of the equation, so I can divide through by m and cancel the m's, and now I have a equals g. The g value, 9.8 meters per second per second, is the acceleration at the very top of the loop. Now I can say a equals v squared over r, and rearranging so I can get v by itself, I'd have v squared equals a times r. Wow, now I know my r is 1.8 meters and my a is 9.8. I can solve for v squared, it comes out to be 17.64 meters squared per second squared. Now if I take the square root of both sides, I get this value for the speed, and it comes out to be 4.2 meters per second.