A 0.527 kg metal cylinder is placed inside the top of a plastic tube, the lower
ID: 1591799 • Letter: A
Question
A 0.527 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 5.71 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 the initial acceleration of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm.
Explanation / Answer
The weight of the cylinder is
(0.527 kg)(9.8 m/s^2) = 5.16 N
The cross-sectional area of the tube is
pi*(0.002685 m)^2 = 2.2648 x 10^(-5) m^2
The "gauge" pressure due to the cylinder's weight is
weight/area = 227.834 kPa.
Add on the atmospheric pressure of 1 atm,
which is 101.325 kPa, and the absolute pressure
when the cylinder comes to rest is 339.3 kPa.
a = [{0.03*(10)**(0.00571²)}-(0.527*9.81)
= -4.86 m/s²
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.