Circular and Satellite Motion Legacy Problem #12 Guided Solution

This problem is about a student who is moving through a looped loop on a roller coaster ride. As with many circular motion problems, my strategy is going to center around the use of a free body diagram. So I construct a diagram representing the top of the loop. It's a part of a circle. So I draw the top part of the circle and I put a dot at the so called 12 o'clock position. I record information right there next to the dot. For instance, that Sheila's mass is 62 kilograms and that the radius at the very top of the loop is 12 meters. What I'm looking for is the V, the speed, at the top of the loop. The last piece of information that I know will be very important. And it's the fact that the normal force experienced by Sheila is one half of her weight. I could write this as an equation. The is represents an equal sign. The normal force could be written as F norm. So I have F norm equal and then the one half of her weight would be 0.5 times her weight. Or 0.5 times mg. Now I turn my attention to the free body diagram. I know at the top of the loop Sheila is upside down in her chair. Thus the normal force is above her, the seat is above her and it's pressing down on her. I have two forces acting on her at the top of the loop and they're both down. Gravity which is always down. And I draw an arrow and I label it F grab. A normal force which is usually up but down in this situation since Sheila is upside down and the seat is above her. So I have two forces down and one of them is gravity and it's value is mg. The other one is F norm and it's value is 0.5 times mg. Now I could calculate the values of these forces but I'm not going to do that. Instead I'm going to say the F net is equal to F grab plus F norm. They're both in the same direction so when you add them up you get F net equal 1.5 times mg. Now I could substitute m values and g values into this equation and get F net but I'm not going to. Instead I'm going to say that this 1.5 times mg, the F net, is equal to ma. And when I do that I'll notice I now have an equation with m's in them, both sides. If I divide through by m's the m's would cancel and I now have an equation that says 1.5 g is equal to a. The value of having this is I can now calculate the acceleration. This is just 1.5 times 9.8 and it comes out to be 14.7 meters per second per second. Ultimately I wish to get the speed and I know that I have an equation that relates a, v, and r. For circular motion that equation is a equals v squared over r. It can be rearranged to say v squared equals a times r. So to solve for v I can substitute values of a, 14.7, and r, 12 meters, into this equation and then take the square root of both sides. When I do I get 13.2816 meters per second. I can round that to two digits. That's 13 meters per second.