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Stokes' law & Terminal Velocity is considered one the most difficult concept.
29 Questions around this concept.
Spherical balls of radius are falling in a viscous fluid of viscosity with a velocity. The retarding viscous force acting on the spherical ball is:
A spherical solid ball of volume V is made of a material of density 1. It is falling through a liquid of density 2 (2 < 1). Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed . The terminal speed of the ball is
If the terminal speed of a sphere of gold is 0.2 m/s in a viscous liquid , find the terminal speed (in m/sec) of a sphere of silver of the same size in the same liquid.
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When a body moves through a fluid then The fluid exerts a viscous force on the body to oppose its motion.
And according to Stokes' law, the magnitude of the viscous force depends on the shape and size of the body, its speed and the viscosity of the fluid.
So for the below figure
If a sphere of radius r moves with velocity v through a fluid of viscosity ,
Then using Stokes' law the viscous force (F) opposing the motion of the sphere is given by
$
F=6 \pi \eta r v
$
Where
$\eta$ - coefficient viscosity
$r$ - radius
$v-v e l o c i t y$
When the spherical body is dropped in a viscous fluid, it is first accelerated and then it's acceleration becomes zero and it attains a constant velocity and this constant velocity is known as terminal velocity.
For a spherical body of radius r is dropped in a viscous fluid, The forces acting on it are shown in the below figure.
So Forces acting on the body are
1. Weight of Body (W)
$
W=m g=\frac{4}{3} \pi r^3 \rho g
$
Where $\rho \rightarrow$ density of body
2. Buoyant/ Thrust Force (T of $F_B$ )
$
T=F_B=\frac{4}{3} \pi r^3 \sigma g
$
where $\sigma \rightarrow$ density of fluid
3. Viscous force (F)
$
F=6 \pi \eta r v
$
So when the body attains terminal velocity the net force acting on the body is zero.
Apply force balance
$
\begin{aligned}
& F_B+F=W \\
& \rightarrow 6 \pi \eta r v+\frac{4}{3} \pi r^3 \sigma g=\frac{4}{3} \pi r^3 \rho g \\
& \rightarrow 6 \pi \eta r v=\frac{4}{3} \pi r^3 g(\rho-\sigma) \\
& \rightarrow v_t=\frac{2}{9} \frac{r^2(\rho-\sigma)}{\eta} g
\end{aligned}
$
Where $v_T=$ terminal velocity
From this formula, we can say that
I.e Spherical body attains constant velocity in a downward direction.
I.e Spherical body attains constant velocity in an upward direction.
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