terminal speed of a sphere falling a viscous fluid?

Question

Derive an expression for the terminal speed vt of a sphere falling in a viscous fluid in terms of the sphere's radius r and density p and the fluid density n, assuming that the flow is laminar so that Stoke's law is valid. Hint: Use FBD

Answer 1

♦ what is this mysterious FBD? Still I continue; Balance of forces
0=f1+f2+f3, where weight the ball f1=m*g, m=(4/3)*pi r^3 *p is its mass, Archimedes force f2=-(4/3)*pi r^3 *n*g, force on the Stox’s ball f3=6pi*ηur, η is viscosity, u is speed in question (your vt); thus
♠ 6pi*ηur = (4/3)*pi r^3 *(p –n)*g, hence
u = (2/9)* r^2 *(p –n)*g / η;

Answer 2

The terminal speed occurs when the drag force equals gravity force. Look up the formula for drag to get it as a function of velocity. The mass of the sphere is (4/3)*π•r^3*rho, and this time g is the gravity force.

Answer 3

an identical mass has 2 different velocities reckoning on the buoyancy difference, so which you will purely take the version in KE of the two: supply the plastic sphere an arbitrary mass of one million kg KE = one million/2 mv^2 no buoyancy speed --> KE = 0.5 * one million kg ( 0.36 m/s)^2 KE = 0.0648 J buoyancy speed--> KE = 0.5 * 1kg (0.24 m/s)^2 KE = 0.0288 J Now purely divide the version in KE before buoyancy and after: 0.0288 J / 0.0648 J = 0.44444 So the fraction of the load this is buoyancy tension is (40 4.4% - a hundred% ) = fifty 5.fifty 5% of finished tension, or 5/9ths