Q10E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

An object of mass 2 kg is released from rest from a platform 30 m above the water and allowed to fall under the influence of gravity. After the object strikes the water, it begins to sink with gravity pulling down and a buoyancy force pushing up. Assume that the force of gravity is constant, no change in momentum occurs on impact with the water, the buoyancy force is 1/2 the weight (weight = mg), and the force due to air resistance or water resistance is proportional to the velocity, with proportionality constant   b₁ = 10 N-sec/m in the air and   b₂= 100 N-sec/m in the water. Find the equation of motion of the object. What is the velocity of the object 1 min after it is released?

Step-by-Step Solution

Answer

• The equation of motion of the object in air is  ${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{962}\mathbf{t}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{3924}\mathbf{(}{\mathbf{e}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\mathbf{1}\mathbf{)}$
• The equation of motion of the object in water is  $$\begin{gathered} \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\left\{\mathbf{1}\mathbf{.}\mathbf{962}\mathbf{t}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{3924}\mathbf{(}{\mathbf{e}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\mathbf{1}\mathbf{)} \\ \mathbf{0}\mathbf{.}\mathbf{0981}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{5}\mathbf{)}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{03728}{\mathbf{e}}^{\mathbf{-}\mathbf{50}\mathbf{t}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{15}\mathbf{.}\mathbf{5}\mathbf{)}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{03728} \\ \right. \end{gathered}$$
• The velocity of the object in air is 1.962 m/sec. 
• The velocity in the water is 0.0981 m/sec.

Step1: Find the equation of motion in air

Here F = 2g-10v

 $$\begin{gathered} \mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=2\mathrm{g}-10\mathrm{v} \\ 2\frac{\mathrm{dv}}{\mathrm{dt}}=2\mathrm{g}-10\mathrm{v} \\ \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{g}-5\mathrm{v} \end{gathered}$$

 $$\begin{gathered} \frac{1}{5}\ln \left|5\mathrm{v}-\mathrm{g}\right|=-\mathrm{t}+{\mathrm{c}}_1 \\ \ln \left|5\mathrm{v}-\mathrm{g}\right|=-5\mathrm{t}+5{\mathrm{c}}_2 \\ \ln \left|5\mathrm{v}-\mathrm{g}\right|=-5\mathrm{t}+5{\mathrm{c}}_2 \\ 5\mathrm{v}=\mathrm{g}+{\mathrm{c}}_3{\mathrm{e}}^{-5\mathrm{t}} \\ {\mathrm{v}}_1=\frac{1}{5}\mathrm{g}+{\mathrm{ce}}^{-5\mathrm{t}} \end{gathered}$$ (by solving variable separating and integrating)

When v(0) = 0 , c = -1.962

${\mathrm{v}}_1=-1.962-1.962{\mathrm{e}}^{-5\mathrm{t}}······\left(1\right)$    (by putting the value of g)

 ${\mathrm{x}}_1\left(\mathrm{t}\right)=1.962\mathrm{t}+0.3924{\mathrm{e}}^{-5\mathrm{t}}+\mathrm{C}$    (By integrating equation (1) on both sides)

 

When x=0, C=-0.3924

 ${\mathrm{x}}_1\left(\mathrm{t}\right)=1.962\mathrm{t}+0.3924({\mathrm{e}}^{-5\mathrm{t}}-1)$

Hence the equation of motion of the object in air ${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{962}\mathbf{t}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{3924}\mathbf{(}{\mathbf{e}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\mathbf{1}\mathbf{)}$  

Step 2: Find the time and velocity in air

When x₁(t) = 30 then

 $$\begin{gathered} 30=1.962\mathrm{t}+0.3924({\mathrm{e}}^{-5\mathrm{t}}-1) \\ \mathrm{t}=\frac{30+0.3924}{1.962} \\ \mathrm{t}=15.5 \sec \end{gathered}$$

When t = 15.5 sec then

$$\begin{gathered} {\mathrm{v}}_{1_{}}=1.962-1.962{\mathrm{e}}^{-77.5} \\ =1.962\mathrm{m}/\sec \end{gathered}$$  

Hence the velocity of the object in air is 1.962 m/sec.

Step 3: Find the equation of motion in the water

Here F₂ = -100v and additional buoyancy force F = 1/2 mg

    $$\begin{gathered} \mathrm{F}=2\mathrm{g}-100\mathrm{v}-\frac{2\mathrm{g}}{2} \\ \mathrm{F}=\mathrm{g}-100\mathrm{v} \\ So, \mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{g}-100\mathrm{v} \\ 2\frac{\mathrm{dv}}{\mathrm{dt}}=9.81-100\mathrm{v} \\ \frac{\mathrm{dv}}{\mathrm{dt}}=4.905-50\mathrm{v} \end{gathered}$$   

 $\frac{1}{50}\ln \left|50\mathrm{v}-4.905\right|=-\mathrm{t}+{\mathrm{c}}_1$ (by solving variable separating and integrating)

$$\begin{gathered} 50\mathrm{v}-4.905={\mathrm{e}}^{-50\mathrm{t}+\mathrm{c}} \\ {\mathrm{v}}_2=0.0981+{\mathrm{Ce}}^{-50\mathrm{t}} \end{gathered}$$ 

When v₂ (0) = 1.962, C = 1.864

${\mathrm{v}}_2=0.0981+1.864{\mathrm{e}}^{-50\mathrm{t}}$ 

 ${\mathrm{x}}_2\left(\mathrm{t}\right)=0.0981\mathrm{t}-0.03728{\mathrm{e}}^{-50\mathrm{t}}+\mathrm{C}$

When x₂(0) = 0 ,C = 0.03728

 ${\mathrm{x}}_2\left(\mathrm{t}\right)=0.0981\mathrm{t}-0.03728{\mathrm{e}}^{-50\mathrm{t}}+0.03728$

On combining the both equations 

  $$\begin{gathered} \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\left\{\mathbf{1}\mathbf{.}\mathbf{962}\mathbf{t}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{3924}\mathbf{(}{\mathbf{e}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\mathbf{1}\mathbf{)} \\ \mathbf{0}\mathbf{.}\mathbf{0981}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{5}\mathbf{)}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{03728}{\mathbf{e}}^{\mathbf{-}\mathbf{50}\mathbf{t}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{15}\mathbf{.}\mathbf{5}\mathbf{)}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{03728} \\ \right. \end{gathered}$$

Hence, the equation of motion of the object in water is

$$\begin{gathered} \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\{1.962t+0.3924(e^{-5t}-1) \\ 0.0981(t-5)-0.03728e^{-50t(t-15.5)}+0.03728 \end{gathered}$$

Step 4: Find the velocity in water

Now t after 1 min t = 60-15.5 = 45.5 sec.

 $$\begin{gathered} {\mathrm{v}}_2(45.5)=0.0981+1.864{\mathrm{e}}^{-50t} \\ =0.0981 \mathrm{m}/\sec \end{gathered}$$

Hence, the velocity in the water is 0.0981 m/sec.