Unformatted text preview: the pendulum is conserved. E i = E f ⇒ (1/2)mv 2 + mgL(1 – cos θ ) = 0 + mgL v = √ (2gL cos θ ) = 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 θ ) = (1/2)mv 2 + mg(2L) (1/2)mv 2 + mgL(1 – cos θ ) = (1/2)mgL + mg(2L) v = √ (3gL + 2gL cos θ ) = 7.94 m/s...
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 Summer '08
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 Force, Work, General Relativity, Gravitational Force, Tomoyuki Nakayama, workenergy theorem yields

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