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    dks2114's Avatar
    dks2114 Posts: 32, Reputation: 2
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    #1

    Oct 8, 2007, 05:13 PM
    Conservation of Energy_ Correct?
    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
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Oct 9, 2007, 08:54 AM
    Close - however, the potential energy that the block has to overcome to get to the top of the loop is 2*mgr.
    dks2114's Avatar
    dks2114 Posts: 32, Reputation: 2
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    #3

    Oct 9, 2007, 10:07 AM
    Why is it: 2*mgr.
    Where does the 2 come from?
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Oct 9, 2007, 12:07 PM
    Quote Originally Posted by dks2114
    Why is it: 2*mgr.
    Where does the 2 come from?
    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:
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    dks2114 Posts: 32, Reputation: 2
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    #5

    Oct 9, 2007, 12:18 PM
    Ohhh wow thank you that actually makes sense
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