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 $\left({\rho }_{s}=densityofthesphere,r=radius\right)$.

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Solution

## 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:We know, the downward force on a body when it is placed in a liquid is ${F}_{d}=Bodyweight-buoyancyforceonthebody$.So, the downward force exerted on the body is ${F}_{d}=\left(\frac{4}{3}{\mathrm{\pi r}}^{3}×{\mathrm{\rho }}_{\mathrm{s}}\right)×g-\left(\frac{4}{3}{\mathrm{\pi r}}^{3}×{\rho }_{w}\right)×g$, 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:The terminal velocity is the constant velocity when a body freely falls through a liquid.The terminal velocity is defined by the form, ${v}_{t}=\frac{2{r}^{2}\left({\rho }_{s}-{\rho }_{w}\right)g}{9\eta }$, 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${F}_{d}=\left(\frac{4}{3}{\mathrm{\pi r}}^{3}×{\mathrm{\rho }}_{\mathrm{s}}\right)×g-\left(\frac{4}{3}{\mathrm{\pi r}}^{3}×{\rho }_{w}\right)×g$So, ${F}_{d}=\frac{4}{3}{\mathrm{\pi r}}^{3}g\left[{\mathrm{\rho }}_{\mathrm{s}}-{\rho }_{w}\right].................\left(1\right)\phantom{\rule{0ex}{0ex}}$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, $a=\frac{F}{m}=\frac{\frac{4}{3}{\mathrm{\pi r}}^{3}g\left({\mathrm{\rho }}_{\mathrm{s}}-{\rho }_{w}\right)}{\frac{4}{3}{\mathrm{\pi r}}^{3}×{\rho }_{s}}=\frac{\left({\mathrm{\rho }}_{\mathrm{s}}-{\rho }_{w}\right)g}{{\rho }_{s}}\phantom{\rule{0ex}{0ex}}⇒a=\frac{\left({\mathrm{\rho }}_{\mathrm{s}}-{\rho }_{w}\right)g}{{\rho }_{s}}......................\left(2\right)$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, ${v}_{t}=0+\frac{a}{2}t;\left(\mathrm{sin}ce,{a}_{avg}=\frac{a}{2}\right)\phantom{\rule{0ex}{0ex}}⇒t=\frac{2{v}_{t}}{a}=\frac{\frac{2×2{r}^{2}\left({\mathrm{\rho }}_{\mathrm{s}}-{\rho }_{w}\right)g}{9\eta }}{\frac{\left({\mathrm{\rho }}_{\mathrm{s}}-{\rho }_{w}\right)g}{{\rho }_{s}}}=\frac{4{\rho }_{s}{r}^{2}}{9\eta }\left(\mathrm{sin}ce,terminalvelocity,{v}_{t}=\frac{2{r}^{2}\left({\rho }_{s}-{\rho }_{w}\right)g}{9\eta }\right)\phantom{\rule{0ex}{0ex}}ort=\frac{4{\rho }_{s}{r}^{2}}{9\eta }.$Therefore, the time to attain the terminal velocity is $\frac{4{\rho }_{s}{r}^{2}}{9\eta }$.

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