ANY HELP IS MUCH APPRECIATED, I ALWAYS RATE, THANKS! . A particle descending thr

Question

ANY HELP IS MUCH APPRECIATED, I ALWAYS RATE, THANKS! . A particle descending through a viscous medium experiences africtional drag proportional to its speed s and equal to6Rs, where R is its radius and (eta) isthe viscosity of the medium. If the acceleration of free fall isdenoted g, what will be the terminal velocity of a sphere of radiusR and mass density (rho)? . ** I know that sTerminal = Force / s, but I am notquite sure how to factor in a sphere and mass denisty into this?Thanks! . ANY HELP IS MUCH APPRECIATED, I ALWAYS RATE, THANKS! . A particle descending through a viscous medium experiences africtional drag proportional to its speed s and equal to6Rs, where R is its radius and (eta) isthe viscosity of the medium. If the acceleration of free fall isdenoted g, what will be the terminal velocity of a sphere of radiusR and mass density (rho)? . ** I know that sTerminal = Force / s, but I am notquite sure how to factor in a sphere and mass denisty into this?Thanks! .

Explanation / Answer

Let = density of fluid Force = mass*acceleration = -m*g + *g*V + 6R*v =dv/dt Note gV is the buoyant force. buoyant and drag force act inopposite direction of the gravitational force At terminal velocity, dv/dt = 0 -m*g + *g*V + 6R*v = 0 g*(V - m) + 6R*v = 0 V = (4/3)R3 m = *(4/3)R3 Sub into equation g*(V - m) + 6R*v = 0 g*( -)*[4R3/3] + 6R*v =0 g*( -)*[2R2/3] + 3*v = 0 v = g*( - )*2R2/9 (I notice you didn't discuss the fluid density; have you notdiscussed buoyant force?)

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