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- Thread starter Sirsh
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PhanthomJay

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Therefore mv^2r/r = 5mg - mg

So, mv^2/r = 4mg, mass cancles.

v^2/r = 4g, r = v^2/4g

so (800/3.6)^2 / (4*9.8) = r , so r = ~1260m. ??

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my bad _wolfgang_

thats for bottom ;D

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PhanthomJay

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You are correct that the net centripetal force at the bottom is 4mg, and the centripetal acceleration at the at the bottom is 4g. Now we already established (given) that v = 720 km/hr = 200m/s. So what's that 800/3.6 for the v value?

Therefore mv^2r/r = 5mg - mg

So, mv^2/r = 4mg, mass cancles.

v^2/r = 4g, r = v^2/4g

so (800/3.6)^2 / (4*9.8) = r , so r = ~1260m. ??

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- #32

PhanthomJay

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Now you've got it! I'll leave it up you as an academic exercise if you what to assume something to see what may be happening at the top of the loop. Note that in more typical loop the loop problems, like yoyos in a vertcal circle , or loop the loop roller coasters , the speed is far from constant, so don't take this problem as a typical example of vertcal circular motion problems.

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