Question: A small metal cylinder of mass 0.1 kg is attached to a string and hangs vertically in air. The cylinder is immersed in a container filled with an oil of viscosity 0.2 kg/m·s. The length of the cylinder is 0.5 m and its diameter is 0.05 m.
(a) Calculate the resistance experienced by the cylinder as it falls through the oil. (b) If the cylinder starts from rest at the surface of the oil and falls for 2 seconds, determine its terminal velocity.
(c) Assuming the cylinder continues to fall with a constant velocity after reaching terminal velocity, calculate the power dissipated by the resistance force as the cylinder settles down at terminal velocity.
Given: Mass of the cylinder, m = 0.1 kg Length of the cylinder, L = 0.5 m Diameter of the cylinder, d = 0.05 m Viscosity of the oil, η = 0.2 kg/m·s Acceleration due to gravity, g = 9.8 m/s²
Constants: Density of the oil, ρ = 900 kg/m³ Drag coefficient, C = 0.8
Assume the cylinder's motion is vertical and no other forces are acting on it.
(a) Resistance Calculation:
Step 1: Calculate the cross-sectional area of the cylinder. The cross-sectional area, A = πr² = π(0.025)² = 0.00196 m²
Step 2: Calculate the velocity of the cylinder as it falls through the oil. Using the equation of motion, v = u + at, where:
Substituting the values, we find v = 0 + 9.8 * 2 = 19.6 m/s.
Step 3: Calculate the Reynolds number for the cylinder's motion through the oil. Reynolds number, Re = (ρvd) / η, where:
Substituting the values, we find Re = (900 * 19.6 * 0.05) / 0.2 = 2214
Step 4: Determine the resistance experienced by the cylinder using the drag coefficient equation. Resistance force, F = 0.5 * C * ρ * A * v²
Substituting the values calculated above, we have F = 0.5 * 0.8 * 900 * 0.00196 * (19.6)² = 14.92 N
Therefore, the resistance experienced by the cylinder as it falls through the oil is approximately 14.92 N.
(b) Terminal Velocity Calculation:
Terminal velocity is the point at which the resistance force equals the weight of the cylinder. At terminal velocity, the net force on the cylinder becomes zero.
Resistance force, F = 14.92 N (from part a) Weight of the cylinder, W = mg = 0.1 * 9.8 = 0.98 N
Setting these forces equal, we have:
14.92 N = 0.98 N 0.5 * C * ρ * A * v_term² = mg
Solving for v_term, we find:
v_term² = (2 * mg) / (C * ρ * A) v_term = √[((2 * 0.1 * 9.8) / (0.5 * 0.8 * 900 * 0.00196))] = 8.85 m/s
Therefore, the terminal velocity of the cylinder is approximately 8.85 m/s.
(c) Power Dissipated Calculation:
When the cylinder reaches terminal velocity, it continues to fall with a constant velocity. In this case, the resistance force does work against the motion of the cylinder, which dissipates energy.
Power dissipated, P = F * v_term Substituting the values, we have P = 14.92 * 8.85 = 131.79 W
Therefore, the power dissipated by the resistance force as the cylinder settles down at terminal velocity is approximately 131.79 Watts.