The figure below shows Atwood\'s machine, in which two containersare connected b

Question

The figure below shows Atwood's machine, in which two containersare connected by a cord (of negligible mass) passing over africtionless pulley (also of negligible mass). At time t =0, container 1 has mass 1.30 kg and container 2 has mass3.25 kg, but container 1 is losing mass(through a leak) at the constant rate of 0.170 kg/s. (a) At what rate is the acceleration magnitudeof the containers changing at t = 0?
_______ m/s3

(b) At what rate is the acceleration magnitude of the containerschanging at t = 3.00 s?
_________ m/s3

(c) When does the acceleration reach its maximum value?
(a) At what rate is the acceleration magnitudeof the containers changing at t = 0?
_______ m/s3

(b) At what rate is the acceleration magnitude of the containerschanging at t = 3.00 s?
_________ m/s3

(c) When does the acceleration reach its maximum value?

Explanation / Answer

This is a tricky question. First we start with the expressionfor the acceleration of an Atwoods machine: .     a = g (m2 -m1 )/ (m1 + m2 ) . We can drop in a few numbers for the constant values you aregiven: .     a = 9.80 (3.25 - m) / ( m + 3.25) . . Now... we need to find   da/dt. We can do this bytaking the derivative of the function with respect to time. Weget: .     da/dt = (d/dt) [  9.80(3.25 - m ) ( m + 3.25  )-1   ] .    da/dt =   9.80 [ -1 * (3.25- m) ( m + 3.25   )-2   -   ( m + 3.25   )-1  ] (dm/dt)      where I used the chainrule . Now... simpliy this a bit: .         da/dt  = - 9.80 * dm/dt [ (3.25 - m) + (m +3.25) ] / ( m + 3.25  )2     .      da/dt = -9.80 * dm/dt *6.50 / ( m + 3.25   )2    =     -63.7 * dm/dt / ( m +3.25   )2   . And we are told   dm/dt = -0.170 so .    da/dt = -63.7 * -0.170 / ( m +3.25   )2   =     10.829 / ( m +3.25   )2   . NOW... we can answer some questions... . At t = 0     m = 1.30  and .     da/dt = 10.829 / (1.30 +3.25)2=   0.5231m/s3 . At   t = 3     .           m = 1.30   - 0.17*3  =    0.79 kg   and .     da/dt = 10.829 / (0.79 +3.25)2 =    0.6906 m/s3

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