Question

1.

a)Calculate the terminal velocity of a wooden ball, an iron ball,

and a lead ball -- each of which is 4 inches in diameter.

b).How long (in seconds) can each fall from

rest before air resistance will cause a 5% change in the ball’s

acceleration? How far (in meters) can each fall before air

resistance will cause a 5% change in the ball’s acceleration

Answer 1

The terminal speed is defined

$v=\sqrt{\frac{mg}{b}}$

is called coefficient of friction of the medium which is defined like

$b=\tfrac{1}{2}\,C_{d}\,A\,\rho$

if the medium is the air

$b=\tfrac{1}{2}\,C_{d}\,A\,\rho_{\mathrm{air}}$

where

• $C_{d}$   is called drag coefficient. For spheres value of this coefficient is
•   is the transverse surface of the object. The transverse surface of the spheres is

​ $A=\pi R^{2}$

• $\rho_{\mathrm{air}}$   is the density of the air. Its value is $\mathrm{1.14\,\tfrac{Kg}{m^{3}}}$

then,

$v=\sqrt{\frac{mg}{\tfrac{1}{2}\,C_{d}\,A\,\rho_{\mathrm{air}}}}$

$v=\sqrt{\frac{2\,mg}{\,C_{d}\,\pi\,R^{2}\,\rho_{\mathrm{air}}}}$

Since we don't know the mass of the balls we must rewrite the mass in terms of the density of each ball. So, from the definition of density

$\rho=\frac{m}{V}$

the masses of the spheres are determined by the expression

$m=\rho\,V$

$m=\rho\left( \tfrac{4}{3}\pi R^{3} \right )$

$m=\tfrac{4}{3}\pi R^{3}\rho$

Hence, the terminal speed

$v=\sqrt{\frac{2\,mg}{\,C_{d}\,\pi\,R^{2}\,\rho_{\mathrm{air}}}}$

$v=\sqrt{\frac{2\left( \tfrac{4}{3}\pi R^{3}\rho \right)g}{\,C_{d}\,\pi\,R^{2}\,\rho_{\mathrm{air}}}}$

$v=\sqrt{\frac{8\pi R^{3}\rho\,g}{3\,C_{d}\,\pi\,R^{2}\,\rho_{\mathrm{air}}}}$

$v=\sqrt{\frac{8\,R\,\rho\,g}{3\,C_{d}\,\rho_{\mathrm{air}}}}$

a) The terminal speed of the wooden ball

$v=\sqrt{\frac{8\,R\,\rho_{\mathrm{wooden}}\,g}{3\,C_{d}\,\rho_{\mathrm{air}}}}$

$v=\mathrm{\sqrt{\frac{8\left( 5.08\times10^{-2}\,m \right)\left( 700\,\tfrac{Kg}{m^{3}} \right)\left( 9.81\,\tfrac{m}{s^{2}} \right )}{3\left( 0.47 \right)\left( 1.14\,\tfrac{Kg}{m^{3}} \right)}}}$

$v=\mathrm{41.7\,\tfrac{m}{s^{2}}}$

The terminal speed of the iron ball

$v=\sqrt{\frac{8\,R\,\rho_{\mathrm{wooden}}\,g}{3\,C_{d}\,\rho_{\mathrm{air}}}}$

$v=\mathrm{\sqrt{\frac{8\left( 5.08\times10^{-2}\,m \right)\left( 7870\,\tfrac{Kg}{m^{3}} \right)\left( 9.81\,\tfrac{m}{s^{2}} \right )}{3\left( 0.47 \right)\left( 1.14\,\tfrac{Kg}{m^{3}} \right)}}}$

$v=\mathrm{139\,\tfrac{m}{s^{2}}}$

The terminal speed of the lead ball

$v=\sqrt{\frac{8\,R\,\rho_{\mathrm{wooden}}\,g}{3\,C_{d}\,\rho_{\mathrm{air}}}}$

$v=\mathrm{\sqrt{\frac{8\left( 5.08\times10^{-2}\,m \right)\left( 1.13\times10^{4}\,\tfrac{Kg}{m^{3}} \right)\left( 9.81\,\tfrac{m}{s^{2}} \right )}{3\left( 0.47 \right)\left( 1.14\,\tfrac{Kg}{m^{3}} \right)}}}$

$v=\mathrm{167\,\tfrac{m}{s^{2}}}$

b) The wooden ball

$v_{\mathrm{y}}=g\,t \begin{matrix} && --> && \end{matrix} t=\frac{v_{\mathrm{y}}}{g}$

$t=\mathrm{\frac{41.7\,\tfrac{m}{s}}{9.81\,\tfrac{m}{s^{2}}}}$

$t=\mathrm{4.25\,s}$

and

$y=\tfrac{1}{2}\,g\,t^{2}$

$y=\mathrm{\tfrac{1}{2}\left( 9.81\,\tfrac{m}{s^{2}} \right)\left( 4.25\,s \right)^{2}}$

$y=\mathrm{88.6\,m}$

The Iron ball

$v_{\mathrm{y}}=g\,t \begin{matrix} && --> && \end{matrix} t=\frac{v_{\mathrm{y}}}{g}$

$t=\mathrm{\frac{139\,\tfrac{m}{s}}{9.81\,\tfrac{m}{s^{2}}}}$

$t=\mathrm{14.2\,s}$

and

$y=\tfrac{1}{2}\,g\,t^{2}$

$y=\mathrm{\tfrac{1}{2}\left( 9.81\,\tfrac{m}{s^{2}} \right)\left( 14.2\,s \right)^{2}}$

$y=\mathrm{989\,m}$

The lead ball

$v_{\mathrm{y}}=g\,t \begin{matrix} && --> && \end{matrix} t=\frac{v_{\mathrm{y}}}{g}$

$t=\mathrm{\frac{167\,\tfrac{m}{s}}{9.81\,\tfrac{m}{s^{2}}}}$

$t=\mathrm{17.0\,s}$

and

$y=\tfrac{1}{2}\,g\,t^{2}$

$y=\mathrm{\tfrac{1}{2}\left( 9.81\,\tfrac{m}{s^{2}} \right)\left( 17.0\,s \right)^{2}}$

$y=\mathrm{1417\,m}$