Question
What is terminal velocity and derive an expression for it?
✓
Answer
Terminal velocity is the maximum constant velocity acquired by the body which is falling freely in a viscous medium, due to the balanced net downward force acting on the body with the upward resistive viscous force offered by the medium on the body. When a small spherical body falls freely through a viscous medium then 3 forces act on it:-
i. Weight of body acting vertically downwards.
ii. Up thrust due to buoyancy = weight of fluid displaced by the body, acting upwards.
iii. Viscous drag $\left( F _{ V }\right)$ or resistive viscous force acting in the direction opposite to the motion of body.
Let $\rho=$ Density of the material of the spherical body
$r =$ Radius of the spherical body
$\sigma=$ Density of the viscous medium.
$\therefore$ True weight of the body $= W =$ volume of the body $\times$ density of the body $\times g$
$
\therefore W=\frac{4}{3} \pi r^3 \rho g
$
Up ward thrust by the fluid, $F _{ T }=$ weight of medium displaced by the spherical body $=$ volume of the body $\times$ density of the viscous
$
\begin{aligned}
& \text { medium } \times g \\
& =\frac{4}{3} \pi r^3 \sigma g
\end{aligned}
$
Say, $v_T=$ Terminal velocity of body
According to Stoke's law, viscous drag or viscous force, $F_V=6 \pi \eta v_T$ ( $\eta$ being coefficient of viscosity of the medium)
When the body attains terminal velocity $v_T$, then
$
\begin{aligned}
& F_{T}+F_{V}=W \\
& \Rightarrow \frac{4}{3} \pi r^3 \sigma g+6 \pi \eta r v_T=\frac{4}{3} \pi r^3 \rho g \\
& \therefore v_T=\frac{2 r^2(\rho-\sigma) g}{9 \eta}
\end{aligned}
$
i. $v_T$ directly depends on radius of body and difference of the pressure of material and medium.
ii. $v_T$ inversely depends of co-efficient of viscosity of the medium.
i. Weight of body acting vertically downwards.
ii. Up thrust due to buoyancy = weight of fluid displaced by the body, acting upwards.
iii. Viscous drag $\left( F _{ V }\right)$ or resistive viscous force acting in the direction opposite to the motion of body.
Let $\rho=$ Density of the material of the spherical body
$r =$ Radius of the spherical body
$\sigma=$ Density of the viscous medium.
$\therefore$ True weight of the body $= W =$ volume of the body $\times$ density of the body $\times g$
$
\therefore W=\frac{4}{3} \pi r^3 \rho g
$
Up ward thrust by the fluid, $F _{ T }=$ weight of medium displaced by the spherical body $=$ volume of the body $\times$ density of the viscous
$
\begin{aligned}
& \text { medium } \times g \\
& =\frac{4}{3} \pi r^3 \sigma g
\end{aligned}
$
Say, $v_T=$ Terminal velocity of body
According to Stoke's law, viscous drag or viscous force, $F_V=6 \pi \eta v_T$ ( $\eta$ being coefficient of viscosity of the medium)
When the body attains terminal velocity $v_T$, then
$
\begin{aligned}
& F_{T}+F_{V}=W \\
& \Rightarrow \frac{4}{3} \pi r^3 \sigma g+6 \pi \eta r v_T=\frac{4}{3} \pi r^3 \rho g \\
& \therefore v_T=\frac{2 r^2(\rho-\sigma) g}{9 \eta}
\end{aligned}
$
i. $v_T$ directly depends on radius of body and difference of the pressure of material and medium.
ii. $v_T$ inversely depends of co-efficient of viscosity of the medium.
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