Question
Derive an expression for terminal velocity of a spherical object falling under gravity through a viscous medium
✓
Answer
i.Consider a sphere of the radius (r) and density (ρ) falling under gravity through a liquid of density (σ) and coefficient of viscosity (η) as shown in the figure.
ii.Forces acting on the sphere during downward motion are
a. Viscous force =$ F_v$ = 6πηrv (directed upwards)
b. Weight of the sphere, $(F_g)$
\(mg =\frac{4}{3} \pi r^3 \rho g\) (directed downwards)
c. Upward thrust as Buoyant force $(F_u)$
\(F _{ U }=\frac{4}{3} \pi r^3 \sigma g\) (directed upwards)
iii.As the downward velocity increases, the viscous force increases. A stage is reached when the sphere attains terminal velocity.
iv.When the sphere attains the terminal velocity, the total downward force acting on the sphere is balanced by the total upward force acting on the sphere.
∴ Total downward force = Total upward force
∴ Weight of sphere (mg) = Viscous Force +
Buoyant force due to medium
\(\therefore \frac{4}{3} \pi r^3 \rho g=6 \pi \eta r \vee+\frac{4}{3} \pi r^3 \sigma g\)
\(\therefore 6 \pi \eta r v=\left(\frac{4}{3} \pi r^3 \rho g\right)-\left(\frac{4}{3} \pi r^3 \sigma g\right)\)
\(\therefore 6 \pi \eta r v=\left(\frac{4}{3}\right) \pi r^3 g(\rho-\sigma)\)
\(\therefore v =\left(\frac{4}{3}\right) \pi r^3 g(\rho-\sigma) \times \frac{1}{6 \pi \eta r}\)
\(\therefore v=\frac{2}{9} \frac{r^2 g(\rho-\sigma)}{\eta}\)...(1)
v.This is the expression for the terminal velocity of the sphere falling under gravity through a viscous medium.
ii.Forces acting on the sphere during downward motion are
a. Viscous force =$ F_v$ = 6πηrv (directed upwards)
b. Weight of the sphere, $(F_g)$
\(mg =\frac{4}{3} \pi r^3 \rho g\) (directed downwards)
c. Upward thrust as Buoyant force $(F_u)$
\(F _{ U }=\frac{4}{3} \pi r^3 \sigma g\) (directed upwards)
iii.As the downward velocity increases, the viscous force increases. A stage is reached when the sphere attains terminal velocity.
iv.When the sphere attains the terminal velocity, the total downward force acting on the sphere is balanced by the total upward force acting on the sphere.
∴ Total downward force = Total upward force
∴ Weight of sphere (mg) = Viscous Force +
Buoyant force due to medium
\(\therefore \frac{4}{3} \pi r^3 \rho g=6 \pi \eta r \vee+\frac{4}{3} \pi r^3 \sigma g\)
\(\therefore 6 \pi \eta r v=\left(\frac{4}{3} \pi r^3 \rho g\right)-\left(\frac{4}{3} \pi r^3 \sigma g\right)\)
\(\therefore 6 \pi \eta r v=\left(\frac{4}{3}\right) \pi r^3 g(\rho-\sigma)\)
\(\therefore v =\left(\frac{4}{3}\right) \pi r^3 g(\rho-\sigma) \times \frac{1}{6 \pi \eta r}\)
\(\therefore v=\frac{2}{9} \frac{r^2 g(\rho-\sigma)}{\eta}\)...(1)
v.This is the expression for the terminal velocity of the sphere falling under gravity through a viscous medium.
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