Q 3.4-1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

An object of mass 5 kg is released from rest 1000 m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant b = 50 N-sec/m, determine the equation of motion of the object. When will the object strike the ground?

Step-by-Step Solution

Answer

• The equation of motion of the object is $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{981}\mathbf{t}\mathbf{-}\mathbf{0}.0\mathbf{981}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{10}\mathbf{t}}\mathbf{)} \text{m}$

 

• The time taken by the object is 1019 sec.

Step 1: Find the equation of motion of an object

The given values are   ${\text{m = 5, g = 9.81, v}}_{\text{0}}\text{ = 0, b = 50,}$

 

The equation of motion is    $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{mgt}}{\mathbf{b}}\mathbf{+}\frac{\mathbf{m}}{\mathbf{b}}\left(\mathbf{v}\mathbf{-}\frac{\mathbf{mg}}{\mathbf{b}}\right)\left(\mathbf{1}\mathbf{-}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{bt}}{\mathbf{m}}}\right)$

 

Put all the given values 

 $$\begin{gathered} \frac{5(9.81)\mathrm{t}}{50}+\frac{5}{50}(0-\frac{5(9.81)}{50}\left)\right(1-{\mathrm{e}}^{\frac{-50\mathrm{t}}{5}}) \\ \frac{(9.81)\mathrm{t}}{10}-\frac{(9.81)}{100}\left)\right(1-{\mathrm{e}}^{-10\mathrm{t}}) \\ \mathrm{x}\left(\mathrm{t}\right)=\frac{(9.81)\mathrm{t}}{10}-\frac{(9.81)}{100}\left)\right(1-{\mathrm{e}}^{-10\mathrm{t}}) \\ \mathrm{x}\left(\mathrm{t}\right)=0.981\mathrm{t}-0.0981(1-{\mathrm{e}}^{-10\mathrm{t}})\mathrm{m} \end{gathered}$$

 Hence the equation of motion is   $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{981}\mathbf{t}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{981}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{10}\mathbf{t}}\mathbf{)}\mathbf{ }\mathbf{\text{m}}$

Step 2: Finding when will the object strike the ground

$$\begin{gathered} 1000=0.981\mathrm{t}-0.0981(1-{\mathrm{e}}^{-10\mathrm{t}}) \\ 1000=0.981\mathrm{t}-0.0981 (Because value of is so small so neglecting) \\ 1000+0.0981=0.981\mathrm{t} \\ 1000.0981=0.981\mathrm{t} \\ \mathrm{t}=1000.0981/0.981 \\ \mathrm{t}=1019 \sec \end{gathered}$$

Hence, the object will strike the ground in 1019 sec.