Question

A 0.389 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 7.41 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 1

The initial acceleration is 97.96 m/s²

Step-by-step explanation:

Given data:

m = mass of metal cylinder = 0.389 kg

r = radius of plastic tube = 7.41 mm = 7.41x10⁻³m

P₀ = pressure = 1 atm = 1.013x10⁵Pa

The pressure inside the tube is:

`P=P_(0) +(mg)/(\pi r^(2) ) =1.013x10^(5) +(0.389*9.8)/(\pi (7.41x10^(-3))^(2) ) =1.234x10^(5) Pa`

The pressure applied by the plunger is:

`P_(p) =2.79*P=2.79*1.234x10^(5) =3.443x10^(5) Pa`

The difference between the pressure applied by the plunger and the pressure inside the tube is:

`delt-P=3.443x10^(5) -1.234x10^(5) =2.209x10^(5) Pa`

The initial acceleration is:

`a=(delta-P*A)/(m) =(2.209x10^(5)*\pi *(7.41x10^(-3))^(2) )/(0.389) =97.96m/s^(2)`