The figure below shows Atwood\'s machine, in which two containers are connected

Question

The figure below shows Atwood's machine, in which two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time t = 0, container 1 has mass 1.30 kg and container 2 has mass 2.70 kg, but container 1 is losing mass (through a leak) at the constant rate of 0.240 kg/s.

a) At what rate is the acceleration magnitude of the containers changing at t = 0?
(b) At what rate is the acceleration magnitude of the containers changing at t = 3.00 s?
(c) When does the acceleration reach its maximum value?

Explanation / Answer

Data: Mass, m1 = 1.3 kg Mass, m2 = 2.7 kg Rate of loss of mass, dm1 / dt = - 0.240 kg/s Solution: Acceleration, a = [ ( m2 - m2 ) / ( m2 + m1 ) ] g (a) Rate of change of acceleration: da / dt = [ ( da / dm1) ( dm1 / dt ) ]            = [ - 2 m2 g / ( m2 + m1 )^2 ] * ( dm1 / dt )            = [ - 2 * 2.7 * 9.8 / ( 2.7 + 1.3 )^2 ] * - 0.240            = 0.79 m/s^2 Ans: Rate of change of acceleration = 0.79 m/s^2 Ans: Rate of change of acceleration = 0.79 m/s^2 (b) t = 3 s m1' = m1 + ( dm1 /dt) t        = 1.3 + (-0.240) * 3        = 1.3 - 0.72        = 0.58 kg Rate of change of acceleration: da / dt = [ ( da / dm1) ( dm1 / dt ) ]            = [ - 2 m2 g / ( m2 + m1' )^2 ] * ( dm1 / dt )            = [ - 2 * 2.7 * 9.8 / ( 2.7 + 0.58 )^2 ] * - 0.240            = 1.18 m/s^2 Ans: Rate of change of acceleration = 1.18 m/s^2 (c) The acceleration becomes when,                0 = m1' = m1 + (dm1/dt) t                            0 = 1.3 + (- 0.24) t                             t = 5.42 s Ans: t = 5.42 s Rate of change of acceleration: da / dt = [ ( da / dm1) ( dm1 / dt ) ]            = [ - 2 m2 g / ( m2 + m1' )^2 ] * ( dm1 / dt )            = [ - 2 * 2.7 * 9.8 / ( 2.7 + 0.58 )^2 ] * - 0.240            = 1.18 m/s^2 Ans: Rate of change of acceleration = 1.18 m/s^2 (c) The acceleration becomes when,                0 = m1' = m1 + (dm1/dt) t                            0 = 1.3 + (- 0.24) t                             t = 5.42 s Ans: t = 5.42 s

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