qz6sol_3885f10 - the pendulum is conserved. E i = E f...

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TA: Tomoyuki Nakayama Monday, October 11, 2010 PHY 2048: Physic 1, Discussion Section 3885 Quiz 6 (Homework Set #7) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ The figure below right shows a pendulum of length L = 1.50 m. Its bob (which effectively has all the mass) has speed v 0 when the cord makes an angle of θ 0 = 50.0 with the vertical. a) What is the least value that v 0 can have if the pendulum is to swing down and then up to a horizontal position? Two forces are exerted on the pendulum bob: tension and the gravitational force. Only the gravitational force does work. Therefore, the mechanical energy of
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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 work-energy 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.

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