Solved Problem on Dynamics
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A particle of mass m is launched vertically upwards with initial velocity v0 and rises under the action of a resistance force proportional to the speed. Determine:
a) The equation of velocity as a function of time;
b) The equation of displacement as a function of time;
c) the maximum height reached by the particle.
Problem data:
- Mass of particle: m;
- Initial velocity of particle: v0;
- Drag coefficient: b.
We choose a reference frame pointing in an upward direction from the point where the particle is
launched. The gravitational force Fg and the force of resistance FR
oppose the motion of the particle (Figure 1).
Figure 1
Solution
a) Applying Newton's Second Law to the particle
\[ \bbox[#99CCFF,10px]
{\mathbf{F}=m\dot{{\mathbf{v}}}}
\]
\[
-{\mathbf{F}_{g}}-{\mathbf{F}}_{R}=m{\mathbf{\dot{v}}}
\]
the gravitational force is given by
\[ \bbox[#99CCFF,10px]
{\mathbf{F}_{g}=m\mathbf{g}}
\]
the force of resistance is given by
\[ \bbox[#99CCFF,10px]
{\mathbf{F}_{R}=b\mathbf{v}}
\]
substituting these expressions
\[
-mg\;\mathbf{j}-bv\;\mathbf{j}=m\dot{v}\;\mathbf{j}
\]
as there is motion only in one dimension we have the magnitude
\[
-mg-bv=m\dot{v}
\]
writing
\( \dot{v}=\frac{dv}{dt} \)
and separating the variables
\[
\begin{gather}
-mg-bv=m\frac{dv}{dt}\\
m\frac{dv}{dt}=-b\left(\frac{mg}{b}+v\right)\\
\frac{dv}{\left(\dfrac{mg}{b}+v\right)}=-{\frac{b}{m}}dt
\end{gather}
\]
integrating the speed on the left-hand side from v0 to v(t), the speed at
any time t, and on the right-hand side integrating the time from 0 to t
\[
\begin{gather}
\int_{v_{0}}^{{v(t)}}\frac{dv}{\left(\dfrac{mg}{b}+v\right)}=\int_{0}^{t}-\frac{b}{m}dt\\
\int_{v_{0}}^{{v(t)}}\frac{dv}{\left(\dfrac{mg}{b}+v\right)}=-{\frac{b}{m}}\int_{0}^{t}dt
\end{gather}
\]
Integration of \( \displaystyle \int_{v_{0}}^{{v(t)}}\frac{dv}{\left(\dfrac{mg}{b}+v\right)} \)
Changing the variable
for \( v=v_{0} \)
we have \( u=\dfrac{mg}{b}+v_{0} \)
for \( v=v(t) \)
we have \( u=\dfrac{mg}{b}+v(t) \)
substituting in the integral
Changing the variable
\[
\begin{align}
& u=\frac{mg}{b}+v\\
& \frac{du}{dv}=0+1\Rightarrow du=dv
\end{align}
\]
changing the limits of integration
for \( v=v_{0} \)
we have \( u=\dfrac{mg}{b}+v_{0} \)
for \( v=v(t) \)
we have \( u=\dfrac{mg}{b}+v(t) \)
substituting in the integral
\[
\begin{split}
\int_{\frac{{mg}}{b}+v_{0}}^{\frac{{mg}}{b}+v(t)}\frac{du}{u} &=\left.\ln u\right|_{\frac{{mg}}{b}+v_{0}}^{\frac{{mg}}{b}+v(t)}=\ln\left(\frac{mg}{b}+v(t)\right)-\ln\left(\frac{mg}{b}+v_{0}\right)=\\[5pt]
&=\ln\left(\frac{\dfrac{mg}{b}+v(t)}{\dfrac{mg}{b}+v_{0}}\right)=\ln\left(\frac{\dfrac{mg+bv(t)}{\cancel{b}}}{\dfrac{mg+bv_{0}}{\cancel{b}}}\right)=\\[5pt]
&=\ln\left(\frac{mg+bv(t)}{mg+bv_{0}}\right)
\end{split}
\]
Integration of \( \displaystyle \int_{0}^{t}dt \)
\[
\int_{0}^{t}dt=\left.t\;\right|_{\;0}^{\;t}=t-0=t
\]
\[
\begin{gather}
\ln\left(\frac{mg+bv(t)}{mg+bv_{0}}\right)=-{\frac{b}{m}}t\\[5pt]
\frac{mg+bv(t)}{mg+bv_{0}}=\operatorname{e}^{-{\frac{b}{m}}t}\\[5pt]
mg+bv(t)=(mg+bv_{0})\;\operatorname{e}^{-{\frac{b}{m}}t}\\[5pt]
bv(t)=(mg+bv_{0})\;\operatorname{e}^{-{\frac{b}{m}}t}-mg
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{v(t)=\frac{1}{b}(mg+bv_{0})\;\operatorname{e}^{-{\frac{b}{m}}t}-\frac{mg}{b}}
\]
b) Substituting \( \dot{x}=\frac{dx}{dt} \), in the equation found above
\[
\begin{gather}
\frac{dx}{dt}=\frac{1}{b}(mg+bv_{0})\;\operatorname{e}^{-{\frac{b}{m}}t}-\frac{mg}{b}\\
dx=\left[\frac{1}{b}(mg+bv_{0})\;\operatorname{e}^{-{\frac{b}{m}}t}-\frac{mg}{b}\right]dt
\end{gather}
\]
Integrating the position on the left-hand side from 0 to x(t), the position at any time
t, and on the right-hand side integrating the time from 0 to t. The integral of difference of
two functions is the difference of their integrals
\[
\int_{0}^{{x(t)}}dx=\int_{0}^{t}\frac{1}{b}(mg+bv_{0})\;\operatorname{e}^{-{\frac{b}{m}}t}dt-\int_{0}^{t}\frac{mg}{b}dt
\]
In the first integral the term
\( \dfrac{1}{b}(mg+bv_{0}) \),
and in the second integral the term
\( \dfrac{mg}{b} \),
are constant, and they are moved outside of the integral
\[
\int_{0}^{{x(t)}}dx=\frac{1}{b}(mg+bv_{0})\int_{0}^{t}\;\operatorname{e}^{-{\frac{b}{m}}t}dt-\frac{mg}{b}\int_{0}^{t}dt
\]
Integration of \( \displaystyle \int_{0}^{{x(t)}}dx \)
\[
\int_{0}^{{x(t)}}dx=\left.x\;\right|_{\;0}^{\;x(t)}=x(t)-0=x(t)
\]
Integration of \( \displaystyle \int_{0}^{t}\;\operatorname{e}^{-{\frac{b}{m}}t}dt \)
Changing the variable
for \( t=0 \)
we have \( u=-{\dfrac{b}{m}}.0=0 \)
for \( t=t \)
we have \( u=-{\dfrac{b}{m}}t \)
substituting in the integral
Changing the variable
\[
\begin{align}
& u=-{\frac{b}{m}}t\\
& \frac{du}{dt}=-{\frac{b}{m}}\Rightarrow dt=-{\frac{m}{b}}du
\end{align}
\]
changing the limits of integration
for \( t=0 \)
we have \( u=-{\dfrac{b}{m}}.0=0 \)
for \( t=t \)
we have \( u=-{\dfrac{b}{m}}t \)
substituting in the integral
\[
\begin{split}
\int_{0}^{-{\frac{b}{m}}t}\;\operatorname{e}^{u}\left(-{\frac{m}{b}}\right)du &=-{\frac{m}{b}}\int_{0}^{-{\frac{b}{m}}t}\;\operatorname{e}^{u}du=-{\frac{m}{b}}\left.\;\operatorname{e}^{u}\;\right|_{0}^{-{\frac{b}{m}}t}=\\[5pt]
&=-\frac{m}{b}\left(\operatorname{e}^{-{\frac{b}{m}}t}-\operatorname{e}^{0}\right)=-{\frac{m}{b}}\left(\operatorname{e}^{-{\frac{b}{m}}t}-1\right)
\end{split}
\]
The integral in dt has already been calculated above.
\[
x(t)=\frac{1}{b}(mg+bv_{0})\left[-{\frac{m}{b}}\left(\operatorname{e}^{-{\frac{b}{m}}t}-1\right)\right]-\frac{mg}{b}t
\]
\[ \bbox[#FFCCCC,10px]
{x(t)=-{\frac{m}{b^{2}}}(mg+bv_{0})\left[\operatorname{e}^{-{\frac{b}{m}}t}-1\right]-\frac{mg}{b}t}
\]
c) When the particle reaches the maximum height, its speed is equal to zero, v(t) = 0, substituting this value in the expression of the item (a), we get the climb time of the particle
\[
\begin{gather}
0=\frac{1}{b}(mg+bv_{0})\;\operatorname{e}^{-{\frac{b}{m}}t}-\frac{mg}{b}\\[5pt]
\frac{1}{b}(mg+bv_{0})\;\operatorname{e}^{-{\frac{b}{m}}t}=\frac{mg}{b}\\[5pt]
\operatorname{e}^{-{\frac{b}{m}}t}=\frac{mg}{mg+bv_{0}}\\[5pt]
-{\frac{b}{m}}t=\ln\left(\frac{mg}{mg+bv_{0}}\right)\\[5pt]
t=-{\frac{m}{b}}\ln\left(\frac{mg}{mg+bv_{0}}\right)
\end{gather}
\]
substituting this value in the expression of item (b), we have the maximum height of the particle
\[
\begin{gather}
x(t)=-{\frac{m}{b^{2}}}(mg+bv_{0})\left[\exp\left(-{\frac{b}{m}}\left(-{\frac{m}{b}}\ln\left(\frac{mg}{mg+bv_{0}}\right)\right)\right)-1\right]-\frac{mg}{b}\left(-{\frac{m}{b}}\ln\left(\frac{mg}{mg+bv_{0}}\right)\right)\\[5pt]
x(t)=-{\frac{m}{b^{2}}}(mg+bv_{0})\left[\exp\left(\frac{b}{m}\frac{m}{b}\ln\left(\frac{mg}{mg+bv_{0}}\right)\right)-1\right]+\frac{mg}{b}\frac{m}{b}\ln\left(\frac{mg}{mg+bv_{0}}\right)\\[5pt]
x(t)=-{\frac{m}{b^{2}}}(mg+bv_{0})\left[\exp\left(\ln\left(\frac{mg}{mg+bv_{0}}\right)\right)-1\right]+\frac{m^{2}g}{b^{2}}\ln\left(\frac{mg}{mg+bv_{0}}\right)\\[5pt]
x(t)=-{\frac{m}{b^{2}}}(mg+bv_{0})\left[\frac{mg}{mg+bv_{0}}-1\right]+\frac{m^{2}g}{b^{2}}\ln\left(\frac{mg}{mg+bv_{0}}\right)\\[5pt]
x(t)=-{\frac{m}{b^{2}}}(mg+bv_{0})\left[\frac{mg-(mg+bv_{0})}{mg+bv_{0}}\right]+\frac{m^{2}g}{b^{2}}\ln\left(\frac{mg}{mg+bv_{0}}\right)\\[5pt]
x(t)=-{\frac{m}{b^{2}}}(mg+bv_{0})\left[\frac{mg-mg-bv_{0}}{mg+bv_{0}}\right]+\frac{m^{2}g}{b^{2}}\ln\left(\frac{mg}{mg+bv_{0}}\right)\\[5pt]
x(t)=-{\frac{m}{b^{2}}}(mg+bv_{0})\left[\frac{-bv_{0}}{mg+bv_{0}}\right]+\frac{m^{2}g}{b^{2}}\ln\left(\frac{mg}{mg+bv_{0}}\right)\\[5pt]
x(t)=\frac{mv_{0}}{b}+\frac{m^{2}g}{b^{2}}\ln\left(\frac{mg}{mg+bv_{0}}\right)
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{x(t)=\frac{m}{b}\left[v_{0}+\frac{mg}{b}\ln\left(\frac{mg}{mg+bv_{0}}\right)\right]}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .