A spherical solid ball of volume V is made of a material of density 1. It is falling through aliquid of density 2(2 < 1). Assume that the liquid applies a viscous force on the ball that isproportional to the square of its speed v, i.e. Fviscous = kv2(k > 0). The terminal speed of theball is:
Question
A spherical solid ball of volume V is made of a material of density 1. It is falling through aliquid of density 2(2 < 1). Assume that the liquid applies a viscous force on the ball that isproportional to the square of its speed v, i.e. Fviscous = kv2(k > 0). The terminal speed of theball is:
Solution
The terminal speed of the ball is reached when the forces acting on the ball are balanced. These forces are the gravitational force acting downwards and the buoyant and viscous forces acting upwards.
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The gravitational force (F_grav) acting on the ball is given by the equation F_grav = mg, where m is the mass of the ball and g is the acceleration due to gravity. The mass m can be expressed as the density of the ball (ρ1) times its volume (V), so F_grav = ρ1V*g.
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The buoyant force (F_buoy) acting on the ball is given by the equation F_buoy = ρ2Vg, where ρ2 is the density of the liquid.
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The viscous force (F_viscous) acting on the ball is given by the equation F_viscous = k*v^2, where k is a positive constant and v is the speed of the ball.
At terminal speed, the sum of the buoyant and viscous forces equals the gravitational force, so we have the equation ρ1Vg = ρ2Vg + k*v^2.
Solving this equation for v gives the terminal speed of the ball:
v = sqrt((ρ1 - ρ2)Vg / k)
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