9-19-08LectPHY2048

9-19-08LectPHY2048 - More applications of Newtons second...

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F T –T θ m 2 m 1 || 1 m gsin −θ θ m 1 g m 2 g Last class we solved a problem in which we effectively bent our coordinate axis (at a pulley) to describe the motion of the two sides of the system. || More applications of Newton’s second law
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We now bend the coordinate system 180 o to solve the dynamics of what is called Atwood’s Machine. m 1 m 2 massless pulley, rope For the circumstance to the right suppose m 2 > m 1 . We chose this “direction” of acceleration as being positive . When the masses are released m 2 will accelerate downwards, while m 1 will accelerate upwards. a a Direction is in quotes because this means that acceleration upwards on the left is positive , but acceleration downwards on the right is positive. This is required by the constraint of the rope that dictates that the when one accelerates up the other accelerates down.
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m 1 m 2 T –T m 2 g –m 1 g Having chosen the signs of the of acceleration on the two sides also fixes the signs of the forces on the two sides. On the left we have the tension, T , from the rope acting upward as positive and the force of gravity acting downward coming in as negative: –m 1 g (as usual). On the right, however, since acceleration downward is positive, forces downward are positive while those upward are negative. So there we have –T for the upward tension in the rope and +m 2 g .
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21 11 (m m ) Tm g m g −= + And using this in the expression for m 1 allows solving for T , 12 2m m Tg = + Adding the equations eliminates T 2112 mg mg ma ma =+ m )g m )a ag = + For m 1 : 1 FT m g m a = For, m 2 : 22 2 m g m a = −+ = With these definitions Newton’s 2 nd law becomes: m 1 m 2 T –T m 2 g –m 1 g
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21 (m m ) ag = + 12 2m m Tg = + m 1 m 2 T –T m 2 g –m 1 g , Do these make sense? i.e. m 2 accelerates down with a = g. If instead m 2 = 0 , a = – g , i.e. m 1 accelerates down at the right rate & T = 0 . Good. If m 1 = m 2 = m, a = 0 , and T = mg . Also good . If m 1 = 0 , the first eqn. gives, 2 2 0) g == + While the second eqn gives, for m 1 = 0 , T = 0. Both of which would be expected .
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21 (m m ) ag = + m 1 m 2 T –T m 2 g –m 1 g Notice that the acceleration of the two masses is the acceleration of gravity, g, times the difference in the masses divided by their sum.
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9-19-08LectPHY2048 - More applications of Newtons second...

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