A 0.550 kg metal cylinder is placed inside the top of a plastic tube, the lower

Question

A 0.550 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 7.07 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 of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm.

Explanation / Answer

Pressure is given by:

P = Force/Area

Initially pressure will be

Pi = Patm + mg/A

Pf = 2.31*Pi

Pressure difference = 2.31Pi - Pi = 1.31*pi

Net force will be

Fnet = dP*A = 1.31*Pi*A

Fnet = m*a = 1.31*Pi*A

a = 1.31*Pi*A/m

Using given values:

m = 0.55 kg

A = pi*r^2

A = pi*(7.07*10^-3)^2 = 1.57*10^-4 m^2

Pi = 1.01*10^5 + 0.55*9.81/(1.57*10^-4) = 1.354*10^5 Pa

So,

a = 1.31*1.354*10^5*1.57*10^-4/0.550

a = 50.63 m/sec^2

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