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December 2nd, 2007, 03:37 PM   #1
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Physics: Circular Motion

1. A carnival clown rides a motorcycle down a ramp and around a "loop-the-loop." If the loop has the radius of 18m, what is the slowest speed the rider can have at the top of the loop to avoid falling? (Hint: At the slowest speed, at the top of the loop, the track does not exert force on the motorcycle).

I know equations like F=ma, v=(2)(pi)(r) / T, a=v^2 / r. But, I don't know how to solve the physics question above. Can anyone help me on this one a little?

Thanks,

J.
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December 2nd, 2007, 04:29 PM   #2
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In order to go in a perfect circle, the net force on an object must be mv^2/r. That is taken straight from the two equations f=ma, and a=v^2/r.

Now, since at the top of the loop, the normal force is working downward, (with gravity), the minimum velocity is when the normal force equals 0, or the only force on the track is gravity (W = mg)

so, we know that mg = mv^2/r,
so
g = v^2/r

Can you take it form there?
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December 2nd, 2007, 05:28 PM   #3
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So, you're saying that g=v^2 / r, and I'm assuming that g is the acceleration in the earth which is 9.8 m/s^2.

9.8 m/s^2 = v^2 / 18m
356.4 m^2 / s^2 = v^2
v = 19 m/s

Is the final answer 19 m/s correct?
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December 2nd, 2007, 08:12 PM   #4
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Umm... 9.8*18 = 176.4, not 356.4.

Other than that, you did it right.

Do you understand where the g = v^2/r came from, or should I explain a little more?
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December 2nd, 2007, 08:25 PM   #5
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Oops, I did an arithmetic mistake for 9.8 * 18. Thanks for letting me know.

I think I know where g = v^2/r came from. From F=ma, a=v^2/r, and at the earth gravity, F=ma=mg, thus m(v^2/r) = mg, hence the result.

Thanks for the help.
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