Posted by **Jordan** on Friday, October 8, 2010 at 12:07am.

The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of 4.5 m/s, what is the largest value that the radius r can have if the car is to remain in contact with the circular track at all times?

- physics -
**drwls**, Friday, October 8, 2010 at 12:19am
There is no drawing.

Require that the centripetal force at the top of the loop be equal to M g. Solve for the r that satisfies that requirement, with V determined by conservation of energy.

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**jamie**, Friday, October 8, 2010 at 10:21pm
but you don't know hf or ho, how do you you use the conservation of energy equation?

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**JJ**, Monday, October 11, 2010 at 10:29pm
This is what I found....

When the car is at the top of the track the centripetal

force consists of the full weight of the car.

mv2/r = mg

Applying the conservation of energy between the bottom and the top of the track gives

(1/2)mv^2 + mg(2r) = (1/2)mv0^2

Using both of the above equations

v0^2 = 5gr

so

r = v0^2/(5g) = (4.5 m/s)^2/(49.0m/s^2) =

Hope that helps

- physics -
**Anonymous**, Sunday, February 10, 2013 at 2:54pm
jhhhhhhhhhhhbm

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