A 0.596 kg metal cylinder is placed inside the top of a plastic tube, the lower

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

The weight of the metal cylinder is

W = mg

Here, mass of the metal cylinder is m and acceleration due to gravity is g.

The metal cylinder is in equilibrium initially. Its weight is supported by the outside pressure. Weight of the metal cylinder is also equal to pressure times the area. It is

W = PA

Here, initial pressure is P and area of the metal cylinder is A.

When the plunger is suddenly pushed upwards, the pressure between the plunger and the metal cylinder is increased by a factor of 2.55. Thus, the final pressure is

Pf = 2.55 P

Now, the net force exerted on the metal cylinder is

F = -W + Pf A

    = -W + (2.55 P)A

    = -W + 2.55 W

   = 1.55 W

Let a be the initial acceleration of the metal cylinder.

From Newton’s second law of motion, the acceleration on the metal cylinder is equal to net force acting on the metal cylinder divided by the mass of the metal cylinder. It is

A = F / m

   = 1.55 W / m

   = 1.55 mg / m

   = 1.55 g

Substitute 9.8 m/s2 for g in the above equation

a = 1.55 g

    = (1.55) (9.8 m/s2)

    = 15.19 m/s2

Rounding off to three significant figures, the acceleration of the metal cylinder is 15.2 m/s2.

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