Question

A sphere is dropped under gravity through a fluid of viscosity $h$. If the average acceleration is half of the initial acceleration, the time to attain the terminal velocity is $({\rho }_s=densityofthesphere,r=radius)$.

Answer 1

Step 1: Given data:

The viscosity of the fluid is $h$.

The density of the sphere is ${\rho }_s$.

The radius of the sphere is $r$.

The average acceleration is half of the initial acceleration, i.e,$a_{avg}=\frac{a}{2}$, where, a is the initial acceleration of the sphere.

Step 2: The downward force on a body in a liquid:

1. We know, the downward force on a body when it is placed in a liquid is $F_d=Bodyweight-buoyancyforceonthebody$.
2. So, the downward force exerted on the body is $F_d=\left(\frac{4}{3}{\mathrm{\pi r}}^3\times {\mathrm{\rho }}_{\mathrm{s}}\right)\times g-\left(\frac{4}{3}{\mathrm{\pi r}}^3\times {\rho }_w\right)\times g$,
3. where, $\left(\frac{4}{3}{\mathrm{\pi r}}^3\right)$ is the volume of the solid and ${\rho }_w$ is the density of the liquid, $\mathrm{g}$ is the acceleration due to gravity.

Step 3: Terminal velocity:

1. The terminal velocity is the constant velocity when a body freely falls through a liquid.
2. The terminal velocity is defined by the form, $v_t=\frac{2r^2\left({\rho }_s-{\rho }_w\right)g}{9\eta }$,
3. where ${\rho }_w$ is the density of the liquid, ${\rho }_s$ is the density of the body, $\eta$ is the coefficient of viscosity and r is the radius of the solid body.

Step 4: Finding the downward force:

As we know, the downward force on the body is

$F_d=\left(\frac{4}{3}{\mathrm{\pi r}}^3\times {\mathrm{\rho }}_{\mathrm{s}}\right)\times g-\left(\frac{4}{3}{\mathrm{\pi r}}^3\times {\rho }_w\right)\times g$

So, $F_d=\frac{4}{3}{\mathrm{\pi r}}^{3}g\left[{\mathrm{\rho }}_{\mathrm{s}}-{\rho }_w\right].................\left(1\right)$

Step 5: Finding the acceleration:

We know that acceleration, $a=\frac{F}{m}$, where, F is the force on a body and m is the mass of the body.

So,

$$\begin{gathered} a=\frac{F}{m}=\frac{\frac{4}{3}{\mathrm{\pi r}}^{3}g\left({\mathrm{\rho }}_{\mathrm{s}}-{\rho }_w\right)}{\frac{4}{3}{\mathrm{\pi r}}^3\times {\rho }_s}=\frac{\left({\mathrm{\rho }}_{\mathrm{s}}-{\rho }_w\right)g}{{\rho }_s} \\ \Rightarrow a=\frac{\left({\mathrm{\rho }}_{\mathrm{s}}-{\rho }_w\right)g}{{\rho }_s}......................\left(2\right) \end{gathered}$$

Step 6: Finding the time:

Let the time to attain the terminal velocity is $t$.

AS we know, from the kinematics formulae, $v=u+at$, v and u are the final and initial velocity and a is the acceleration of the body.

So,

$$\begin{gathered} v_t=0+\frac{a}{2}t;\left(\sin ce,a_{avg}=\frac{a}{2}\right) \\ \Rightarrow t=\frac{2v_t}{a}=\frac{{\displaystyle \frac{2\times 2r^2\left({\mathrm{\rho }}_{\mathrm{s}}-{\rho }_w\right)g}{9\eta }}}{\frac{\left({\mathrm{\rho }}_{\mathrm{s}}-{\rho }_w\right)g}{{\rho }_s}}=\frac{4{\rho }_s{r}^2}{9\eta }\left(\sin ce,terminalvelocity,v_t=\frac{2r^2\left({\rho }_s-{\rho }_w\right)g}{9\eta }\right) \\ ort=\frac{4{\rho }_s{r}^2}{9\eta }. \end{gathered}$$

Therefore, the time to attain the terminal velocity is $\frac{4{\rho }_s{r}^2}{9\eta }$.