Solved Problem on Classical Mechanics
Português
English
advertisement
A projectile of mass m is fired with an initial velocity of v0 making an angle θ with the horizontal line. In the projectile acts the force of resistance due to air is proportional to speed. Determine the velocity and displacement equations as a function of time.
Problem data:
- Mass of the projectile: m;
- Initial speed of the projectile: v0;
- Angle of shot with the horizontal: θ;
- Drag coefficient: b.
We choose a xy reference frame at the launch point. The initial velocity can be written using the
unit vectors i and j
\[
\mathbf{v}_{0}=v_{0x}\;\mathbf{i}+v_{0y}\;\mathbf{j}
\]
In Figure 2 the initial velocity can be decomposed in the x and y axes
\[
\begin{gather}
\cos \theta =\frac{v_{0x}}{v_{0}}\Rightarrow v_{0x}=v_{0}\cos \theta \tag{I}\\[10pt]
\sin \theta =\frac{v_{0y}}{v_{0}}\Rightarrow v_{0y}=v_{0}\sin \theta \tag{II}
\end{gather}
\]
Figure 2
Solution
Applying Newton's Second Law to the projectile (Figure 1)
\[ \bbox[#99CCFF,10px]
{\mathbf{F}=m\dot{v}}
\]
\[
-{\mathbf{F}_{g}}-{\mathbf{F}}_{R}=m\dot{v}
\]
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
\[
\begin{gather}
-m\mathbf{g}-b\mathbf{v}=m\mathbf{\dot{v}}\\
-mg\;\mathbf{j}-b(v_{x}\;\mathbf{i}+v_{y}\;\mathbf{j})=m({\dot{v}}_{x}\;\mathbf{i}+{\dot{v}}_{y}\;\mathbf{j})\\
-mg\;\mathbf{j}-bv_{x}\;\mathbf{i}-bv_{y}\;\mathbf{j}=m{\dot{v}}_{x}\;\mathbf{i}+m{\dot{v}}_{y}\;\mathbf{j}
\end{gather}
\]
where vx and vy are the velocity components in the directions i
and j.
Separating the components and writing
\( {\dot{v}}_{x}=\frac{dv_{x}}{dt} \):
- Direction i:
\[
\begin{gather}
-bv_{x}=m{\dot{v}}_{x}\\
-bv_{x}=m\frac{dv_{x}}{dt}\\
\frac{dv_{x}}{v_{x}}=-{\frac{b}{m}}dt
\end{gather}
\]
integrating on both sides of the equation, the limits of integration will be from the initial speed in
direction i, v0x, to the speed at any instant vx(t),
and the limits of time will be from the initial instant 0 to t
\[
\begin{gather}
\int_{v_{0x}}^{{v_{x}(t)}}\frac{dv_{x}}{v_{x}}=\int_{0}^{t}-\frac{b}{m}dt\\
\int_{v_{0x}}^{{v_{x}(t)}}\frac{dv_{x}}{v_{x}}=-{\frac{b}{m}}\int_{0}^{t}dt
\end{gather}
\]
Integration of \( \displaystyle \int_{v_{0x}}^{{v_{x}(t)}}\frac{dv_{x}}{v_{x}} \)
\[
\int_{v_{0x}}^{{v_{x}(t)}}\frac{dv_{x}}{v_{x}}=\left.\ln v_{x}\right|_{v_{0x}}^{v_{x}(t)}=\ln v_{x}(t)-\ln v_{0x}=\ln\left(\frac{v_{x}(t)}{v_{0x}}\right)
\]
Integration of \( \displaystyle \int_{0}^{t}dt \)
\[
\int_{0}^{t}dt=\left.t\;\right|_{\;0}^{\;t}=t-0=t
\]
\[
\begin{gather}
\ln\left(\frac{v_{x}(t)}{v_{0x}}\right)=-{\frac{b}{m}}t\\[5pt]
\frac{v_{x}(t)}{v_{0x}}=\operatorname{e}^{-{\frac{b}{m}}t}\\[5pt]
v_{x}(t)=v_{0x}\operatorname{e}^{-{\frac{b}{m}}t}
\end{gather}
\]
substituting the expression (I) to the initial velocity v0x
\[
\begin{gather}
v_{x}(t)=v_{0}\cos \theta \;\operatorname{e}^{-{\frac{b}{m}}t} \tag{III}
\end{gather}
\]
Using
\( v_{x}=\frac{dx}{dt} \),
substituting in the expression of the speed above, we find the equation of the displacement
\[
\begin{gather}
\frac{dx}{dt}=v_{0}\cos \theta\;\operatorname{e}^{-{\frac{b}{m}}t}\\
dx=v_{0}\cos \theta\;\operatorname{e}^{-{\frac{b}{m}}t}\;dt
\end{gather}
\]
integrating on both sides of the equation, the limits of integration will be from the starting position 0
to the position at any instant x(t), and the time from the initial instant 0 to any
instant t
\[
\int_{0}^{{x(t)}}dx=\int_{0}^{t}v_{0}\cos \theta\;\operatorname{e}^{-{\frac{b}{m}}t}dt
\]
In the integral of the right-hand side of the equation, the term
\( v_{0}\cos \theta \)
is constant and can be moved outside of the integral
\[
\int_{0}^{{x(t)}}dx=v_{0}\cos \theta \int_{0}^{t}\operatorname{e}^{-{\frac{b}{m}}t}dt
\]
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 \( \displaystyle \int_{0}^{t}\;\operatorname{e}^{-{\frac{b}{m}}t}dt \)
making the change of variable
for \( t=0 \)
we have \( u=-{\frac{b}{m}}.0=0 \)
for \( t=t \)
we have \( u=-{\frac{b}{m}}t \)
substituting in the integral
making the change of variable
\[
\begin{align}
& u=-{\frac{b}{m}}t\\
& \frac{du}{dt}=-{\frac{b}{m}}\Rightarrow dt=-{\frac{m}{b}}du
\end{align}
\]
making the change in the limits of integration
for \( t=0 \)
we have \( u=-{\frac{b}{m}}.0=0 \)
for \( t=t \)
we have \( u=-{\frac{b}{m}}t \)
substituting in the integral
\[
\begin{split}
\int_{0}^{-{\frac{b}{m}}t}\;\operatorname{e}^{u}\left(-{\frac{m}{b}}\right)du &=-{\frac{m}{b}}\int_{0}^{-{\frac{b}{m}}t}\;\operatorname{e}^{u}du=-{\frac{m}{b}}\left.\;\operatorname{e}^{u}\;\right|_{0}^{-{\frac{b}{m}}t}=\\[5pt]
&=-\frac{m}{b}\left(\operatorname{e}^{-{\frac{b}{m}}t}-\operatorname{e}^{0}\right)=-{\frac{m}{b}}\left(\operatorname{e}^{-{\frac{b}{m}}t}-1\right)
\end{split}
\]
\[
\begin{gather}
x(t)=v_{0}\cos \theta\left[-{\frac{m}{b}}\left(\operatorname{e}^{-{\frac{b}{m}}t}-1\right)\right]\\[5pt]
x(t)=\frac{m}{b}v_{0}\cos \theta\left(1-\operatorname{e}^{-{\frac{b}{m}}t}\right) \tag{IV}
\end{gather}
\]
- Direction j:
\[
\begin{gather}
-mg-bv_{y}=m{\dot{v}}_{y}\\[5pt]
-mg-bv_{y}=m\frac{dv_{y}}{dt}\\[5pt]
-b\left(\frac{mg}{b}+v_{y}\right)=m\frac{dv_{y}}{dt}\\[5pt]
\frac{dv_{y}}{\frac{mg}{b}+v_{y}}=-{\frac{b}{m}}dt
\end{gather}
\]
integrating on both sides of the equation, the limits of integration will be from the initial speed
v0y to the speed at any instant vy(t), and the time from
the initial instant 0 to any instant t
\[
\begin{gather}
\int_{v_{0y}}^{{v_{y}(t)}}\frac{dv_{y}}{\left(\frac{mg}{b}+v_{y}\right)}=\int_{0}^{t}-\frac{b}{m}dt\\[5pt]
\int_{v_{0y}}^{{v_{y}(t)}}\frac{dv_{y}}{\left(\frac{mg}{b}+v_{y}\right)}=-{\frac{b}{m}}\int_{0}^{t}dt
\end{gather}
\]
Integration of \( \displaystyle \int_{v_{0y}}^{{v_{y}(t)}}\frac{dv_{y}}{\left(\frac{mg}{b}+v_{y}\right)} \)
making the change of variable
for \( v_{y}=v_{0y} \)
we have \( u=\frac{mg}{b}+v_{0y} \)
for \( v_{y}=v_{y}(t) \)
we have \( u=\frac{mg}{b}+v_{y}(t) \)
substituting in the integral
making the change of variable
\[
\begin{align}
& u=\frac{mg}{b}+v_{y}\\
& \frac{du}{dv}=0+1\Rightarrow du=dv_{y}
\end{align}
\]
making the change in the extremes of integration
for \( v_{y}=v_{0y} \)
we have \( u=\frac{mg}{b}+v_{0y} \)
for \( v_{y}=v_{y}(t) \)
we have \( u=\frac{mg}{b}+v_{y}(t) \)
substituting in the integral
\[
\begin{split}
\int_{\frac{{mg}}{b}+v_{0y}}^{\frac{{mg}}{b}+v_{y}(t)}\frac{du}{u} &=\left.\ln u\right|_{\frac{{mg}}{b}+v_{0y}}^{\frac{{mg}}{b}+v_{y}(t)}=\ln\left(\frac{mg}{b}+v_{y}(t)\right)-\ln\left(\frac{mg}{b}+v_{0y}\right)=\\[5pt]
&=\ln\left(\frac{\dfrac{mg}{b}+v_{y}(t)}{\dfrac{mg}{b}+v_{0y}}\right)=\ln\left(\frac{\dfrac{mg+bv_{y}(t)}{\cancel{b}}}{\dfrac{mg+bv_{0y}}{\cancel{b}}}\right)=\\[5pt]
&=\ln\left(\frac{mg+bv_{y}(t)}{mg+bv_{0y}}\right)
\end{split}
\]
Integration of \( \displaystyle \int_{0}^{t}dt \)
\[
\int_{0}^{t}dt=\left.t\;\right|_{\;0}^{\;t}=t-0=t
\]
\[
\begin{gather}
\ln\left(\frac{mg+bv_{y}(t)}{mg+bv_{0y}}\right)=-{\frac{b}{m}}t\\[5pt]
\frac{mg+bv_{y}(t)}{mg+bv_{0y}}=\operatorname{e}^{-{\frac{b}{m}}t}\\[5pt]
mg+bv_{y}(t)=(mg+bv_{0y})\;\operatorname{e}^{-{\frac{b}{m}}t}\\[5pt]
bv_{y}(t)=(mg+bv_{0y})\;\operatorname{e}^{-{\frac{b}{m}}t}-mg\\[5pt]
v_{y}(t)=\frac{1}{b}(mg+bv_{0y})\;\operatorname{e}^{-{\frac{b}{m}}t}-\frac{mg}{b}
\end{gather}
\]
substituting the expression (II) to the initial velocity v0y
\[
\begin{gather}
v_{y}(t)=\frac{1}{b}(mg+bv_{0}\sin \theta)\;\operatorname{e}^{-{\frac{b}{m}}t}-\frac{mg}{b} \tag{V}
\end{gather}
\]
Using
\( v_{y}=\frac{dy}{dt} \),
substituting this value in the expression of the speed above, we find the equation of the motion
\[
\begin{gather}
\frac{dy}{dt}=\frac{1}{b}(mg+bv_{0}\sin \theta)\;\operatorname{e}^{-{\frac{b}{m}}t}-\frac{mg}{b}\\[5pt]
dy=\left[\frac{1}{b}(mg+bv_{0}\sin \theta)\;\operatorname{e}^{-{\frac{b}{m}}t}-\frac{mg}{b}\right]dt
\end{gather}
\]
integrating on both sides of the equation, the limits of integration will be from the initial position 0
to the position at any instant y(t), and the time from the initial instant 0, to any instant
t
\[
\int_{0}^{{y(t)}}dy=\int_{0}^{t}\frac{1}{b}(mg+bv_{y}\sin \theta)\;\operatorname{e}^{-{\frac{b}{m}}t}dt-\int_{0}^{t}\frac{mg}{b}dt
\]
In the first integral on the right-hand side, the term
\( \frac{1}{b}(mg+bv_{y}\sin \theta ) \)
and in the second integral the term
\( \frac{mg}{b} \)
are constant and can be moved outside of the integral
\[
\int_{0}^{{y(t)}}dy=\frac{1}{b}(mg+bv_{y}\sin \theta)\int_{0}^{t}\operatorname{e}^{-{\frac{b}{m}}t}dt-\frac{mg}{b}\int_{0}^{t}dt
\]
Integration of \( \displaystyle \int_{0}^{{y(t)}}dy \)
\[
\int_{0}^{{y(t)}}dy=\left.y\;\right|_{\;0}^{\;y(t)}=y(t)-0=y(t)
\]
Integration of \( \displaystyle \int_{0}^{t}\;\operatorname{e}^{-{\frac{b}{m}}t}dt \)
making the change of 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
making the change of variable
\[
\begin{align}
& u=-{\frac{b}{m}}t\\
& \frac{du}{dt}=-{\frac{b}{m}}\Rightarrow dt=-{\frac{m}{b}}du
\end{align}
\]
making the change in the extremes 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 been calculated above.
\[
\begin{gather}
y(t)=\frac{1}{b}(mg+bv_{y}\sin \theta)\left[-{\frac{m}{b}}\left(\operatorname{e}^{-{\frac{b}{m}}t}-1\right)\right]-\frac{mg}{b}t\\[5pt]
y(t)=\frac{m}{b^{2}}(mg+bv_{y}\sin \theta)\left(1-\operatorname{e}^{-{\frac{b}{m}}t}\right)-\frac{mg}{b}t \tag{VI}
\end{gather}
\]
The velocity will be
\[
\mathbf{v}(t)=v_{x}(t)\;\mathbf{i}+v_{y}(t)\;\mathbf{j}
\]
substituting the expressions (III) and (V)
\[ \bbox[#FFCCCC,10px]
{\mathbf{v}(t)=v_{0}\cos \theta\;\operatorname{e}^{-{\frac{b}{m}}t}\;\mathbf{i}+\left[\frac{1}{b}(mg+bv_{0}\sin \theta)\;\operatorname{e}^{-{\frac{b}{m}}t}-\frac{mg}{b}\right]\;\mathbf{j}}
\]
The displacement will be
\[
\mathbf{\text{r}}(t)=x(t)\;\mathbf{i}+y(t)\;\mathbf{j}
\]
substituting the expressions (IV) and (VI)
\[ \bbox[#FFCCCC,10px]
{\mathbf{\text{r}}(t)=\frac{m}{b}v_{0}\cos \theta\left(1-\operatorname{e}^{-{\frac{b}{m}}t}\right)\;\mathbf{i}+\left[\frac{m}{b^{2}}(mg+bv_{y}\sin \theta)\left(1-\operatorname{e}^{-{\frac{b}{m}}t}\right)-\frac{mg}{b}t\right]\;\mathbf{j}}
\]
advertisement
Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .