A 0.343 kg metal cylinder is placed inside the top of a plastic tube, the lower

Question

A 0.343 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 2.55, what is the initial acceleration of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm. y= 0 a= 0 Number 144 m/s2

Explanation / Answer

Solution-

Initially the cylinder is at rest
so the pressure = weight of cylinder / area + atmospleric presusre

= 0.343 * 9.81 / ( pi * 0.00707^2 ) + 101325 = 122763.51 Pa
now after the plunger suddenly moves
new pressure = 2.79 * inital pressure = 2.79 *122763.51

= 245527.02 Pa
Therefore for the cylinder
( new pressure - atmospheric presure ) * area -   weight = mass * accleraion
( 245527.02 - 101325 ) * ( pi * 0.00707^2 )    -   0.343 * 9.81 = 0.343 * a

19.27=0.343 * a

=>so acceleriaon = a

=56.18 m/sec2

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