Question

An object moving in a liquid experiences a linear drag force: ${\overrightarrow{F}}_{\mathrm{drag}}=(bv$, direction opposite to the motion), where $b$is a constant called the drag coefficient. For a sphere of radius $R,$the drag constant can be computed as $b=6\pi \eta R$, where $\eta$is the viscosity of the liquid.

a. Find an algebraic expression for $v_x\left(t\right)$, the $x$-component as a function of time, for a spherical particle of radius $R$and mass $m$that is shot horizontally with initial speed $v_0$through a liquid of viscosity $\eta$.

b. Water at $20°\mathrm{C}$has viscosity $\eta =1.0\times {10}^{-3}\mathrm{N}\mathrm{s}/{\mathrm{m}}^2$. Suppose a $4.0-cm$-diameter, $33g$ball is shot horizontally into a tank of $20°\mathrm{Cl}$water. How long will it take for the horizontal speed to decrease to $50\%$of its initial value?

Short answer

(a) The algebraic expression for liquid velocity is $v\left(t\right)=v_0-\frac{6\pi nRv}{m}t$

(b) It take for the horizontal speed to decreasw the initial value is$1.46min$

Step-by-step solution

Drag Force Experienced By the Sphere (part a)

a) Given:

Drag force experienced by sphere in liquid $F_{drag}=bv$

Radius of the sphere $=R$

$b=6\pi nR$

Initial speed of sphere $v_i=v_o$$v_i=v_o$

We need to determine the velocity of sphere with Time . As there is no other force except drag force in $x$direction hence its velocity will be affected by drag force only. Also the drag force is opposite to the direction of motion, hence it will lead to decrease in speed with time.

To find speed variations with time we proceed as follow:

$ma=-F_{drag}$

$a=\frac{-F_{drag}}{m}$

$F_{drag}=6\pi \eta Rv$

$⟹a=\frac{6\pi \eta Rv}{m}$

Using equation of motion:

$v_f=v_i+a\Delta t$

$v_f=v_o-\frac{6\pi \eta Rv}{m}t$

$\text{Hence}v\left(t\right)=v_o-\frac{6\pi \eta Rv}{m}t$

Time taken by the sphere. (part b)

b) Given

The coefficient of viscosity$\eta =1\times {10}^{-}3\mathrm{Ns}/{\mathrm{m}}^2$

Diameter $d=4\mathrm{cm}$

$R=2\mathrm{cm}=0.02\mathrm{m}$

Mass $m=33\mathrm{g}=0.033\mathrm{Kg}$

To determine the time taken by sphere to reach $50\%$of initial speed.

Using the result of (a)

$v_f=v_o-\frac{6\pi \eta Rv}{m}t$

$0.5v_o=v_o-\frac{6\times \pi \times 1.0\times {10}^{-3}\times 0.02\times 0.5v}{0.033}t$

$f0.5\backslash \mathrm{not}p=f\frac{6\times \pi \times 1.0\times {10}^{-3}\times 0.02\times 0.5\backslash \mathrm{not}x}{0.033}t$

$t=87.57\mathrm{s}=1.46\min$