A block of mass m slides down a frictionless track, then around the inside of a circular loop-the-loop of radius R. What minimum height h
must the block have to make a full run around the loop without falling off? The answer is to be given as a multiple of R.

Is this right?
centripetal acceleration = g (so it doesn't fall off the tracks)
(v^2) / R = g
v = sqr(gR)

... so...

1/2 (sqr(mgR))^2 + mgR = mgh
1/2mgR + mgR = mgh
3/2mgR = mgh
3/2R = h

Close - however, the potential energy that the block has to overcome to get to the top of the loop is 2*mgr.

Why is it: 2*mgr.
Where does the 2 come from?

Quote:

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Originally Posted by dks2114

Why is it: 2*mgr.
Where does the 2 come from?

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Two times the radius is the diameter of the loop - isn't that the distance the mass has to rise once it enters the loop? See the sketch below:

Ohhh wow thank you that actually makes sense