A 0.458 kg metal cylinder is placed inside the top of a plastic tube, the lower

Question

A 0.458 kg metal cylinder is placed inside the top of a plastic tube, the lower end of which is sealed off by an adjustable plunger, and comes to rest some distance above the plunger. The plastic tube has an inner radius of 5.37 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 3.03, what is fire initial acceleration of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm.

Explanation / Answer

Before the sudden push, the net force on the metal cylinder is zero. This means downward force = upward force

p0 A + m g = p A
where p0 is atmospheric pressure, p the pressure of the trapped air, m the mass of the metal cylinder, g the acceleration due to gravity and A the inner cross sectional area of the plastic tube.

If the pressure p is momentarily increased to 3.03 p, then there is a net upward force of magnitude

Fnet_up = 3.03 p0 A - p0 A - m g
= 2.03 p0 A - m g

This net force gives rise to an acceleration that obeys Newton's second law:

m a = Fnet

m a = 2.03 p0 A - m g

a = 2.03 p0 A /m - g

= 2.03 * 1.01*10^5 N/m^2 * ( pi * (5.37*10^-3 m)^2 ) / 0.458 kg - 9.81 m/s^2
= 40.53 - 9.8

=30.73m/s^2

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