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qz4sol_3705s11

# qz4sol_3705s11 - does work Therefore the mechanical energy...

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TA: Tomoyuki Nakayama Monday, January 24, 2011 PHY 2048: Physic 1, Discussion Section 3705 Quiz 1 (Homework Set #2) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ The figure below right shows a pendulum of length L = 1.20 m. Its bob (which effectively has all the mass) has speed v 0 when the cord makes an angle of θ 0 = 45.0 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: the tension force by the string and the gravitational force. Only the gravitational force, which is conservative,
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Unformatted text preview: does work. Therefore, the mechanical energy of the pendulum is conserved. The energy conservation equation leads to E i = E f ⇒ (1/2)mv 2 + mgL(1 – cos θ ) = 0 + mgL v = √ (2gL cos θ ) = 4.08 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 always equal to or larger than 0. If v 0 has the minimum value, then the tension becomes at the top. 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 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 ) = 7.20 m/s...
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