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Unformatted text preview: the pendulum is conserved. E i = E f (1/2)mv 2 + mgL(1 cos ) = 0 + mgL v = (2gL cos ) = 4.35 m/s b) What is the least value that v 0 can have if the pendulum is to swing down and then up to a vertical position with the cord remaining straight? Since the cord is remaining straight, the bob is in a circular motion and the tension in the string is equal to or larger than 0. If v 0 has the minimum value, then the tension at the top is zero. Therefore, the velocity of the bob at the top is m(v 2 /L) = mg v = (gL) The workenergy theorem yields E i = E f (1/2)mv 2 + mgL(1 cos ) = (1/2)mv 2 + mg(2L) (1/2)mv 2 + mgL(1 cos ) = (1/2)mgL + mg(2L) v = (3gL + 2gL cos ) = 7.94 m/s...
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This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.
 Summer '08
 Field
 Work

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