A 0.596 kg metal cylinder is placed inside the top of a plastic tube, the lower
ID: 1479445 • Letter: A
Question
A 0.596 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 6.05 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 2.55, what is the initial acceleration of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm.Explanation / Answer
To determine the acceleration, we need to know the force which pushing the metal cylinder upward.
Pressure = Force ÷ Area, Force = Pressure * Area
Area = * r^2, r = 6.05 mm = 6.05 * 10^-3 m
Area = * (6.05 * 10^-3)^2
Force = Pressure * * (6.05 * 10^-3)^2
As the plunger is pushed upwards, the pressure increases from 1 atm to 2.55 atm
1 atm = 1.013 * 10^5 N/m^2
2.55 atm = 2.58315 * 10^5 N/m^2
Force = 2.58315 * 10^5 * * (6.05 * 10^-3)^2 29.70 N
This is the force which pushing the cylinder upward.
Weight of cylinder = 0.596 * 9.8 = 5.84 N
Net upward force = 29.70 – 5.84 23.86 N
Acceleration = Net force ÷ mass
Acceleration = 23.86 ÷ 0.596 = 40.03 m/s^2
The acceleration is approximately 40.03 m/s^2
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