Question

A 0.366 kg metal cylinder is placed inside the top of a plastic tube, the lower end of which is sealed off by an adjustable plunger. The cylinder comes to rest some distance above the plunger. The plastic tube has an inner radius of 5.88 mm and is frictionless. Neither the plunger nor the metal cylinder allow any air to flow around them. If the plunger is suddenly pushed upwards, increasing the pressure between the plunger and the metal cylinder by a factor of 2.79 , what is the initial acceleration a of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm and that the top of the tube is open to the air.

Answer in m/s2

Answer 1

initial pressure in chamber

p = 1atm + mg/A = 1.01 x 10⁵Pa + 0.366kg x 9.8m/s² / [π(0.00588m)²]= 1.34 x 10⁵Pa

final pressure p' = 2.79 x p = 2.79 x1.34 x 10⁵ Pa = 3.74 x 10⁵ Pa

for an increase of pressure Δp = p' – p =2.4 x 10⁵Pa

which translates to an increase of force

F = Δp x A = 2.4 x 10⁵Pa x π(0.00588m)² = 26.07 N

instantaneous acceleration a = 26.07 N/ 0.366kg = 71.23 m/s²

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