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Lec11 - Physics 2A Olga Dudko UCSD Physics Lecture 11 Today...

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Today: Pulleys. Atwood’s Machine Circular motion. Centripetal force. Physics 2A Olga Dudko UCSD Physics Lecture 11
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Pulley (Atwood’s Machine) Example Example • Two objects with masses 2.00kg (m 1 ) and 6.00kg (m 2 ) are connected by a light string that passes over a frictionless pulley. Determine the acceleration of each mass and the tension in the string. Solution Solution • First, you must de fi ne a coordinate system. • Let’s choose up as positive y.
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Solution Solution Next, draw a force diagram for each mass separately. Forces are already broken up into components. So we should apply Newton’s 2nd Law separately to each object. 6kg mass F tension, string on 6kg mass F gravity, Earth on 6kg mass 2kg mass F tension, string on 2kg mass F gravity, Earth on 2kg mass Atwood’s Machine
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Solution (cont Solution (cont d) d) • Since the string cannot be stretched, a = a 1 = -a 2 . • Also the tensions will have the same magnitude, T • For the 2kg mass (m 1 ) we have: ± F y = m 1 a 1 ± F y = m 2 a 2 T = m 1 a + m 1 g T - m 1 g = m 1 a T - m 2 g = m 2 (-a) T = m 2 g - m 2 a For the 6kg mass (m 2 ) we have: y equate right- hand sides Atwood’s Machine
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Solution (cont Solution (cont d) d) • Set T = T and solve for the resulting acceleration. a = (9.80m/s 2 )(6kg - 2kg)/(6kg + 2kg) m 1 a + m 2 a = m 2 g - m 1 g m 1 a + m 1 g = m 2 g - m 2 a a = (9.80m/s 2 )(4kg)/(8kg) = 4.90m/s 2 Plugging in the values we get: a(m 2 + m 1 ) = g(m 2 - m 1 ) a = g(m 2 - m 1 )/(m 2 + m 1 ) Atwood’s Machine
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Answer Answer
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