Question

A 0.412 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 6.39 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 2.07, what is the initial acceleration of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm.

PLEASE provide detailed steps on how you got each value that is vital to obtaining the answer. Thanks.

Answer 1

Given that,

Mass of the metal cylinder = M = 0.412 Kg ;

inner radius of the plastic tube = ri = 6.39 mm = 6.39 x 10-3 m

Outside Pressure = 1 atm = 1.01 x 10⁵ N/m²

We need to find the intial acceleration of the metal cylinder. Let it be a.

We know that, Pressure = force / Area

P = F(gravity) / Area

later on the pressure has got increased by a factor of 2.07, So the intial value of pressure be 2.07 - 1 = 1.07 . But in order to increase this much, when the mass remains contant, the acceleration must has got changed to:

a = 1.07 x g = 1.07 x 9.81 = 10.5 m/s²

Hence, the intial acceleration = 10.5 m/s².