Question

An aluminum cylinder weighing 30 N, 6 cm in diameter and 40 cm long, is falling concentrically through a long vertical sleeve of diameter 6.04 cm. The clearance is filled with SAE 50 oil at 20 8 C. Estimate the terminal (zero acceleration) fall velocity. Neglect air drag and assume a linear velocity distribution in the oil. Hint: You are given diameters, not radii.

Answer 1

Answer: Velocity terminal = 0.093m/s

Explanation:

1. We start by evaluating the gap distance between the two cylinders as h = R(sleeve) - R(cylinder)

= (0.0604/2 - 0.06/2)m

= 2×10^-4

Surface are of the cylinder in the drop, which is required in order to evaluate the shearing stress can be expressed as A(cylinder) = π.d.L

= (π×0.06×0.4)m²

= 0.075m²

Since the force of the cylinder's weight is going to balance the shearing force on the walls, we can express the next equation and derive terminal velocity from it.

Shearing stress = u×V.terminal/h = 0.86×V/0.0002

= 4300Vterminal

Therefore, Fw = shearing stress × A

30N = 4300Vterminal × 0.075

V. terminal = 30/4300 m.s

V. terminal = 0.093m/s