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Problem

$\bullet$ $\bullet$ A 2.0 kg piece of wood slide…

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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91 Problem 92 Problem 93 Problem 94 Problem 95 Problem 96 Problem 97 Problem 98 Problem 99 Problem 100 Problem 101 Problem 102 Problem 103 Problem 104 Problem 105 Problem 106

Problem 94 Hard Difficulty

$\bullet$ $\bullet$ Riding a loop-the-loop.
A car in an amusement park
ride travels without friction
along the track shown in the
accompanying figure, starting
from rest at point $A$ . If the loop
the car is currently on has a
radius of $20.0 \mathrm{m},$ find the
minimum height $h$ so that the car will not fall off the track
at the top of the circular part of the loop.

Answer

50 $\mathrm{m}$

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Video Transcript

So we have a car which is going through the track and then it's falling a circular loop. So to successfully complete the new, we should know that the centripetal acceleration, which is acting on the car at this point should be equal to the gravitation force protecting on the car. So if we call the centripetal acceleration as and B squared over R, then that should be equal to the preparation force. So from here, we can see that at this point, the velocity he should be equal to. So let's get it off. M Firsts official vehicle tow squared off. G times are where is the radius? So radius that's denoted by big R Right now we need to figure out the height or the minimum the max way. We need to figure out the height for that. The car can successfully complete the look. So in that case, we can use the conservation of energy at this point and that point and then we can solve for the required height. So, conservation of energy at point a key plus que you eh? We don't have any other form of force present here because the track is friction this and at this point will have Can't be a plus newbie. Now hear the initial velocity of the car is zero. So the zero and using that we can see that kind of energy will be zero here. But they're really potential energy because we have a height. So that's m times g times h A. And in here we'll have potential energy and kind of energy as well. So for Connecticut midget will be half off M v b. So let's call the velocity at this point, Phoebe, and we can see that TV is equal to fee, which we figured out in this part. That and then we'll have the potential energy. So we have the hiders hp. So the pension and you will be empty hb right now from here, if we saw for ah, the S O first for we can do is we can replace this guy over here and then we'll solve for H A. So let's first get rid off this EMS. Then finally, we see that h ey will be equal to M G times twice off art plus half off M g big r divided by m times C. It is five by two, Clark. So that means the height must be a tte least. Aah! 2.5 times are and we know that our is 20 meters. So the height H a will be 2.5 times 20 leaders. That is 50 meters action. Thank you.

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Hugh D. Young

College Physics

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