An object moving in a liquid experiences a linear drag force: $Fudrag=(bv,$direction opposite the motion), where $b$is a constant called the drag coefficient. For a sphere of radius $R,$the drag constant can be computed as$b=6phR$, where $h$is the viscosity of the liquid.

a. Use what you've learned in calculus to prove that

$a_x=v_x\frac{d{v}_x}{dx}$

b. Find an algebraic expression for $v_x\left(x\right)$, component of velocity as a function of distance traveled, for a spherical particle of radius $R$and mass $m$that is shot horizontally with initial speed $v_0$through a liquid of viscosity $h.$

c. Water at $20°\mathrm{C}$has viscosity $\eta =1.0\times {10}^{-3}\mathrm{N}\mathrm{s}/{\mathrm{m}}^2$. Suppose a $1.0-cm$-diameter, $1.0g$marble is shot horizontally into a tank of $20°\mathrm{C}$water at $10cm/s.$. How far will it travel before stopping?

Step by step solution

01

Equation of Calculus

a. Using calculus, prove that

$a_x=v_x\frac{\mathrm{d}{v}_x}{\mathrm{d}x}$

Note:

$a_x=\frac{\mathrm{d}{v}_x}{\mathrm{d}t}$

$v_x=\frac{dx}{dt}$

$a_x=v_x\frac{\mathrm{d}{v}_x}{\mathrm{d}x}$

$\frac{\mathrm{d}{v}_x}{\mathrm{d}t}=v_x\frac{\mathrm{d}{v}_x}{\mathrm{d}x}$

$\frac{v_x}{v_x}·\frac{\mathrm{d}{v}_x}{\mathrm{d}t}=v_x\frac{\mathrm{d}{v}_x}{\mathrm{d}x}$

$v_x·\frac{1}{v_x}·\frac{\mathrm{d}{v}_x}{\mathrm{d}t}=v_x\frac{\mathrm{d}{v}_x}{\mathrm{d}x}$

$v_x·\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}t}}·\frac{\mathrm{d}{v}_x}{\mathrm{d}t}=v_x\frac{\mathrm{d}{v}_x}{\mathrm{d}x}$

$v_x·\frac{\mathrm{d}t}{\mathrm{d}x}·\frac{\mathrm{d}{v}_x}{\mathrm{d}t}=v_x\frac{\mathrm{d}{v}_x}{\mathrm{d}x}$

$v_x·\frac{\mathrm{d}{v}_x}{\mathrm{d}x}=v_x\frac{\mathrm{d}{v}_x}{\mathrm{d}x}$

02

Equation of an algebraic expression

b. Find an algebraic expression for$V_x\left(x\right)\text{;}$

$b=6\pi \eta R$

$F_{drag}=-bv$

$m{a}_x=-6\pi \eta R{v}_x$

$a_x=-\frac{6\pi \eta R{v}_x}{m}$

$v_x\frac{\mathrm{d}{v}_x}{\mathrm{d}x}=-\frac{6\pi \eta R{v}_x}{m}$

$\mathrm{d}{v}_x=-\frac{6\pi \eta R}{m}\mathrm{d}x$

${\int }_{v_0}^{v_x}\mathrm{d}{v}_x={\int }_0^x-\frac{6\pi \eta R}{m}\mathrm{d}x$

$\left[v_x-v_0\right]=-\left[\frac{6\pi \eta R}{m}\left(x\right)-\frac{6\pi \eta R}{m}\left(0\right)\right]$

$v_x=v_0-\frac{6\pi \eta R}{m}x$

03

Calculate the distance traveled by sphere

c. Calculate the distance traveled by the sphere before stopping:

$\eta =1.0\times {10}^{-3}\frac{\mathrm{Ns}}{{\mathrm{m}}^2}$

$R=\frac{1.0\mathrm{cm}}{2}\Rightarrow 0.005\mathrm{m}$

$m=1.0\mathrm{g}\Rightarrow 0.001\mathrm{kg}$

$v_0=10\frac{\mathrm{CII}}{\mathrm{s}}\Rightarrow 0.1\frac{\mathrm{II}}{\mathrm{s}}$

$v_f^2=v_i^2+2a\Delta x$

${\left(v_f\right)}^2={\left(v_0\right)}^2+2\left(-\frac{6\pi \eta R{v}_x}{m}\right)\left(\Delta x\right)$

$v_f=0$

${\left(v_0\right)}^2=\left(\frac{12\pi \eta R{v}_x}{m}\right)\left(\Delta x\right)$

${\left(0.1\frac{\mathrm{m}}{\mathrm{s}}\right)}^2=\left(\frac{12\pi \left(1.0\times {10}^{-3}\frac{\mathrm{Ns}}{{\mathrm{m}}^2}\right)(0.005\mathrm{m})\left(0.1\frac{\mathrm{m}}{\mathrm{s}}\right)}{0.001\mathrm{kg}}\right)\left(\Delta x\right)$

$\Delta x=\frac{{\left(0.1\frac{\mathrm{m}}{\mathrm{s}}\right)}^2}{\left(\frac{12\pi \left(1.0\times {10}^{-3}\frac{\mathrm{Ns}}{{\mathrm{m}}^2}\right)(0.005\mathrm{m})\left(0.1\frac{\mathrm{m}}{\mathrm{s}}\right)}{0.001\mathrm{kg}}\right)}$

$\Delta \mathit{x}=0.53\mathrm{m}$

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