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A Ferris wheel 22.0 m in diameter rotates once every 12.5 s (see Fig. 5–9). What is the ratio of a person’s apparent weight to her real weight at

  1. the top, and
  2. the bottom?
Exercise C.

Figure 5-9 Exercise C.

Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014. Reprinted by permission of Pearson Education Inc., New York.
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Quick Answer: 
  1. $0.716$
  2. $1.28$

Giancoli 7th Edition, Chapter 5, Problem 54


Chapter 5, Problem 54 is solved.

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Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. Here's a picture of somebody going around in this Ferris wheel. At the top of this Ferris wheel, there's a normal force pointing upwards but it's gonna be the smaller than the gravity pulling down because the person is accelerating downwards because sometime later they will move to the right and they will go down. And the normal force is the apparent weight and at the bottom of the Ferris wheel, the normal force would be longer than the gravity down because this normal force up is gonna accelerate them upwards as they as their velocity changes direction in the upwards, towards the upwards direction. So, we are solving for this normal force, in other words, and we are gonna divide it by mg to get the ratio of the normal force to the, you know, regular weight. or real weight. The radius of this Ferris wheel is half the diameter so it's 11 meters radius. We'll take upwards to be the positive direction when we write down Newton's second law. And so we have normal force upwards minus gravity equals negative ma c here where I have, where I know that in part (a), the acceleration is downwards and so I'll put a negative sign in front of this term here. And we can solve for normal force, which is the apparent weight, by moving this gravity term to the right side; it becomes positive. Now, centripetal acceleration is v squared over r and v is the circumference of this Ferris wheel, which is 2πr divided by the period or the time to do one full revolution. So, we'll substitute in for that for v and we end up with 4π squared r over T squared, also known as period squared, for the centripetal acceleration. F g is mg and we'll substitute both of these into our Newton's second law now. So we have F g is mg, which we write here, minus m which was copied times the centripetal acceleration, 4π squared r over T squared and now we are trying to find the ratio to the real weight so we divide this whole normal force apparent weight by mg and when you divide the first term by mg, it becomes 1 and then when you divide the second term by mg, the m's cancel leaving us with a g on the bottom and so here's the equation that we'll plug into; we have 1 minus 4π squared times 11 meters divided by 9.8 meters per second squared times 12.5 seconds squared and that gives 0.716 is the ratio of normal force to real weight. And so this says that the normal force is less than the real weight at the top of the Ferris wheel. So, this is when you get the butterflies sort of, semi-free falling type feeling at the top of the Ferris wheel. And in part (b), the normal force is still upwards and gravity is still downwards and so those things have positive and negative signs respectively still but the acceleration is now upwards and so it's positive ma c and we'll move this gravity to the right and it becomes positive and we have the exact same formulas before except for positive sign there, in front of the ma term. And so we have 1 plus all the same stuff as before and that gives 1.28 is the factor by which the normal force is different from the weight, real weight. And so the normal force is greater than F g now; it's 1.28 times the real weight.