Circular and Satellite Motion Legacy Problem #15 Guided Solution

This problem pertains to a car driving through a circular loop-de-loop. Ultimately, they ask us to determine the normal force acting up on the car at the top of the loop. This would be a difficult problem, a very difficult one, if they didn't step us through A, B, and C up towards the point of calculating this normal force. But fortunately, they do, and in Part A, they ask us to use energy conservation and determine the speed of the car at the top of the loop. So, what I understand about this situation is a car is driving into the loop with an entry speed of 22.1 meters per second, and it has a mass of 938 kilograms. This is information for me to calculate the total amount of energy at the bottom of the loop. It's simply in the form of kE. As the car enters the loop and travels through the loop towards the top, it begins to lose this kE and gain pE, and they give me the height of the loop is 14.2 meters, and so I should be able to calculate the pE at the top. Whatever gain in pE, there's an equivalent loss in kE, so by subtracting the pE from the original kE, I would figure out the kE at the top. So, calculating the kE at the bottom is one-half mv squared, entering 938 kilograms at 22.1 meters per second for m and v, I get about 229,064 joules of energy. The pE at the top is simply mgh, where the m is 938 and the h is 14.2. When I do my math there, I get 130,532 joules of pE gained. So if I subtract this figure from the original kE, I get the kE at the top of the loop. It comes out to be about 98,532 joules of energy. Now I want to find the speed at the top, and kE is equal to one-half mv squared. Doing the algebra on that, the v would be equal to the square root of 2 times kE divided by the mass. So I'll multiply the kE by 2, and then I'll divide by the mass of 938. Then I'll take the square root, and I end up with 14.4945 meters per second. I can round that to three digits, it's about 14.5 meters per second. Now I know how fast the car is going at the top of the loop, and in Part B they asked me to calculate the acceleration. For moving in circles, A is a special case of v squared over r. So I can take my v from Part A, and I can square it, and I can divide by the radius of the loop. The radius of the loop is not explicitly stated, but if I'm thinking here, I know that it's 14.2 centimeters tall. Well that's the diameter cut right through the center, not just the height, but also the diameter. So if I half that value, I get 7.1 meters, and that's my radius. So A equals this v squared from Part A, I've got to square it, and then divide by 7.1, I get my acceleration. It comes out to be 29.5901. There's two significant digits here since the radius is expressed with two significant figures. Now in Part C, I'm going to calculate F net. It's simply ma, and then I'm going to use my F net value in a free body diagram to find the normal force. So I have to think of a car at the top of the loop. It's upside down. The track is above it. So the normal force on the car is not the usual up, it's actually down. And there's also a gravity force down. Together these two forces point down and sum up to the net force. So I can calculate the net force and say that it's equal to F grab plus F norm. Knowing F grab, I can solve for F norm. Here we go. F net is ma. Taking my A from Part B, I multiply by 938 kilograms. I get an F net of 27,756 newtons. Then I calculate F grab by going m, 938 times g, 9.8 newtons per kilogram. I get about 9,192. And then I say F norm plus this F grab equals this F net. Subtracting F grab from each side gives me my F norm. It comes out to be 18,563. I should round that to two significant digits. It's about 1.9 times 10 to the 4th newtons.