Chapter 3, Problem 92P

(a)

To determine

The number of times terminal speed greater than for steel sphere that of the wooden sphere.

Answer to Problem 92P

The steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere.

Explanation of Solution

Write the expression for terminal velocity of the object.

  $v_{\text{term}}=\sqrt{\frac{2mg}{\rho A}}$        (I)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $\rho$ is the density of air, and $A$ is the cross section area of the object.

The cross sectional area of the sphere is,

  $\begin{array}{c}A=\pi r^2\\ =\pi {\left(\frac{d}{2}\right)}^2\end{array}$        (II)

Here, $r$ is the radius of the sphere and $d$ is the diameter of the sphere.

Rewrite the above equation for terminal velocity of wooden sphere.

  $v_{\text{term(wood)}}=\sqrt{\frac{2mg}{\rho A_{\text{wood}}}}$        (III)

Here, $A_{\text{wood}}$ is the cross section area of the wooden sphere.

Rewrite the equation (I) for terminal velocity of steel sphere.

  $v_{\text{term(steel)}}=\sqrt{\frac{2mg}{\rho A_{\text{steel}}}}$        (IV)

Here, $A_{\text{steel}}$ is the cross section area of the steel sphere.

Conclusion:

Solve the equation (III) and (IV).

  $\begin{array}{c}\frac{v_{\text{term}\left(\text{wood}\right)}}{v_{\text{term}\left(\text{steel}\right)}}=\frac{\sqrt{\frac{2mg}{\rho A_{\text{wood}}}}}{\sqrt{\frac{2mg}{\rho A_{\text{steel}}}}}\\ =\sqrt{\frac{A_{\text{steel}}}{A_{\text{wood}}}}\end{array}$

Rewrite the above relation by using equation (2).

  $\begin{array}{c}\frac{v_{\text{term}\left(\text{wood}\right)}}{v_{\text{term}\left(\text{steel}\right)}}=\sqrt{\frac{\pi {\left(\frac{d_{\text{steel}}}{2}\right)}^2}{\pi {\left(\frac{d_{\text{wood}}}{2}\right)}^2}}\\ =\sqrt{\frac{{\left(d_{\text{steel}}\right)}^2}{{\left(d_{\text{wood}}\right)}^2}}\\ =\frac{d_{\text{steel}}}{d_{\text{wood}}}\end{array}$

Substitute $40\text{cm}$ for $d_{\text{wood}}$ and $10\text{cm}$ for $d_{\text{steel}}$ in above relation.

  $\begin{array}{c}\frac{v_{\text{term}\left(\text{wood}\right)}}{v_{\text{term}\left(\text{steel}\right)}}=\frac{10\text{cm}}{40\text{cm}}\\ \frac{v_{\text{term}\left(\text{wood}\right)}}{v_{\text{term}\left(\text{steel}\right)}}=\frac{1}{4}\\ v_{\text{term}\left(\text{wood}\right)}=\frac{1}{4}{v}_{\text{term}\left(\text{steel}\right)}\end{array}$

Therefore, the steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere, due to its smaller cross sectional area.

(b)

To determine

The ratio of the terminal speed of the steel sphere to that of wooden sphere.

Answer to Problem 92P

The terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

Explanation of Solution

From part (a),

The ratio of the terminal speed of the steel sphere to that of wooden sphere for different mass of the sphere with same cross sectional area is,

  $\begin{array}{c}\frac{v_{\text{term}\left(\text{wood}\right)}}{v_{\text{term}\left(\text{steel}\right)}}=\frac{\sqrt{\frac{2m_{\text{wood}}g}{\rho A}}}{\sqrt{\frac{2m_{\text{steel}}g}{\rho A}}}\\ =\sqrt{\frac{m_{\text{wood}}}{m_{\text{steel}}}}\end{array}$

Conclusion:

Substitute $40\text{kg}$ for $m_{\text{steel}}$ and $10\text{kg}$ for $m_{\text{wood}}$ in above relation.

  $\begin{array}{c}\frac{v_{\text{term}\left(\text{wood}\right)}}{v_{\text{term}\left(\text{steel}\right)}}=\sqrt{\frac{10\text{kg}}{40\text{kg}}}\\ \frac{v_{\text{term}\left(\text{wood}\right)}}{v_{\text{term}\left(\text{steel}\right)}}=\frac{1}{2}\\ v_{\text{term}\left(\text{wood}\right)}=\frac{1}{2}{v}_{\text{term}\left(\text{steel}\right)}\end{array}$

Therefore, the terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.