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

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 6.90 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 2.31, 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 1

Let

m = 0.366 kg be the mass of the cylinder,

r = 6.9 mm = 0.0069 m be the radius of the tube,

S = πr² be the area of the cylinder exposed to both inside and outside pressure,

p₀ = 1 atm = 101325 kg/ms² be the ambient pressure,

p₁ (unknown) be the pressure inside the tube when the cylinder comes to rest,

p₂ = 2.31p₁ be the pressure right after the cylinder is pushed upward,

g = 9.81 m/s² be gravitational acceleration.

When the cylinder is at rest, it is subject to three forces:

downward force of gravity: mg

upward force of preasure from the tube: p₁S

downward force of preasure from the atmosphere: p₀S.

When the cylinder is moving up, it is similarly subject to three forces:

downward force of gravity: mg

upward force of pressure from the tube: p₂S

downward force of preasure from the atmosphere: p₀S.

When at rest, the cylinder's acceleration is 0, and

when moving up, its acceleration is a.

We can write Newton's second law identity for both,

while considering upward direction as positive and downward direction as negative:

p₁S - p₀S - mg = 0 (1)

p₂S - p₀S - mg = am (2)

From (1)

p₁S = p₀S + mg

From that

p₂S = 2.31p₁S = 2.31(p₀S + mg)

Substitute into (2)

2.31(p₀S + mg) - p₀S - mg = am

Solve:

a = 1.31(p₀S + mg)/m = 1.31(p₀S/m + g)

Substituting actual numbers

a = 1.31×(101325×π×0.0069²/0.366 + 9.81) = 67 m/s²