2001MechSolutions

2001MechSolutions - The quantity 4 mgD is the decrease in...

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2001 AP Physics Mechanics Exam 1) a) [ ] left s m t t v v t v a o f o f avg 10 33 . 0 37 . 0 23 . 0 17 . 0 = - - - = - - = = b) ( 29 ( 29 ( 29 ( 29 s m kg curve under Area dt F p = + + = = = 60 . 0 02 . 0 10 02 . 0 30 2 1 01 . 0 10 c) kg v p m v m p 5 . 1 23 . 0 17 . 0 60 . 0 = - - - = = = d) ( 29 ( 29 ( 29 ( 29 J mv K J mv K f f o o 0217 . 0 17 . 0 5 . 1 2 1 2 1 0397 . 0 23 . 0 5 . 1 2 1 2 1 2 2 2 2 = = = = = = J J J lost heat K 0180 . 0 0217 . 0 0397 . 0 = - = = 2) a) i) gravity provides centripetal force: R GM v R GM v R GmM R mv J J J = = = 2 2 2 ii) velocity = circumference/period: J J GM R GM R R v R T T R v 3 2 4 2 2 2 π = = = = b) ( 29 ( 29 ( 29 m T GM R GM R T J J 8 3 2 2 4 27 11 3 2 2 3 2 10 59 . 1 4 10 55 . 3 10 9 . 1 10 67 . 6 4 4 × = × × × = = = - c) i) elliptical orbit outside circle ii) elliptical orbit inside circle 3) a) ( 29 ( 29 ( 29 2 2 2 2 2 2 mL L m mr I = = = b) equation of motion for 4 m mass: ma T mg = - 4 equation of motion for rod/masses: ( 29 2 2 2 2 2 r a mL T r a mL Tr I = = = α τ + = = - 2 2 2 2 2 2 2 2 2 4 r L r g a ma r a mL mg c) The quantity 4 mgD is the decrease in the gravitational potential energy of the system. Conservation of energy requires that this decrease be accompanied by an equal increase in other forms of energy in the system. Therefore, the kinetic energy increases by 4 mgD . mgD K U system m 4 4 = = - c) As the mass falls, the rod rotates, and the masses and the ends of the strings swing upward.
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Unformatted text preview: The quantity 4 mgD is the decrease in the gravitational potential energy of the system. Conservation of energy requires that this decrease be accompanied by an equal increase in other forms of energy in the system. There would be an increase in the kinetic energy of the system and an increase in the gravitational potential energy of the masses at the end of the strings as they swing upward. Therefore, the kinetic energy increases less than 4 mgD . m system m U K U ∆ + ∆ = ∆-2 4 mgD U mgD U U K m m m system 4 2 4 2 4 < ∆-= ∆-∆-= ∆ P J P J...
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