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. The cylinder 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 2.07, what is the initial acceleration a of the metal cylinder? Assume the pressure outside of the tube is 1.00 atm m/s2

Answer 1

Here we apply Newton's laws of motion and fluid pressure concept.

An free body diagram of cylinder م ونعمات الك w cylinder COAT WE - Here P A is due to pressure on top of cylinder & (PA) e due to pressure between cylinder & plunger A = Area of cross section of cylinder As cylinder comes to rest cinitially -> (P)(A) – Rot-W = o - TP = Pot Wela W ong m = 0.343 kg A=nX2 r= 5.71 mm Po = atmospheric pressure = latu Now P'= (2:07) ☆ PA Here fret = P'A- Pot-we for fret = :07) PA-PA twt) a fret = (2.07) (P+wc/A) A (PA & we) frer = 2.07 PA +2:07 Wc-POA - We - = (1.07) (Pot the

fret = ma - 11.07) (Pat + wc ama => (1607)CPA + mg) ama o a = (1.07)( PA + g) > a=6-07) [C101,325P2) (%) (5.11 /63m (0.343 kg) - > a = (40.06)(1:07) ms? = +2.86 MASZ