Back exam 1 Work

Problem 1:

Relevant Equations:

$$F=ma=m\frac{dv}{dt}$$ $$F(v)=-F_0{e}^{v/B}=m\frac{dv}{dt}$$

This is a velocity dependent force.

1a.

Finding an expression for velocity using separation of variables

$$-F_0{e}^{v/B}=m\frac{dv}{dt}$$ $$dt=\frac{m}{-F_0{e}^{v/B}}dv$$

Integrate both sides

$$t=\frac{m}{-F_0}\int \frac{1}{e^{v/B}}dv$$ $$t=\frac{m}{-F_0}\int e^{-v/B}dv$$ $$t=\frac{m}{-F_0}(-B)e^{-v/B}+c$$ $$t=\frac{m}{B{F}_0}{e}^{-v/B}+c$$

Solve for V, removing constant and will add in $v_0$ at the end

$$\frac{F_{0}t}{mB}-c=e^{-v/B}$$

Take Ln of both side

$$\ln \left(\frac{F_{0}t}{mB}-c\right)=-\frac{v}{B}$$ $$v(t)=-B\ln \left(\frac{F_{0}t}{mB}-c\right)$$

Solving for c:

$$v(0)=-B\ln \left(0-c\right)$$ $$e^{-v_0/B}=-c$$ $$c=-e^{-v_0/B}$$ $$v(t)=-B\ln \left(\frac{F_{0}t}{mB}+e^{-v_0/B}\right)$$

1b.

$$0=-B\ln \left(\frac{F_{0}t}{mB}+e^{-v_0/B}\right)$$

Because $B\ne 0$, we have to find the value of t that makes the argument of the $\ln$ function 1, because that will be equal to 0

$$1=\frac{F_{0}t}{mB}+e^{-v_0/B}$$ $$1-e^{-v_0/B}=\frac{F_{0}t}{mB}$$ $$t=\frac{mB}{F_0}(1-e^{-v_0/B})$$

1c. Find the equation of motion

$$\frac{dx}{dt}=-B\ln \left(\frac{F_{0}t}{mB}+e^{-v_0/B}\right)$$ $$dx=-B\ln \left(\frac{F_{0}t}{mB}+e^{-v_0/B}\right)dt$$ $$\int dx=\int \ln \left(\frac{F_{0}t}{mB}+e^{v_0/B}\right)dt$$ $$x(t)=\frac{\frac{F_{0}t}{mB}+e^{v_0/B}}{\frac{F_0}{mB}}\ln \left(\frac{F_{0}t}{mB}+e^{-v_0/B}\right)-t$$

Problem 2 (special relativity)

Robbers vehicle is traveling at 0.9c. Police vehicle is traveling at 0.75c. Dart traveling at 0.75c relative to police vehicle.

$$v^{\prime }=\frac{v_1-v_2}{1-\frac{v_1{v}_2}{c^2}}$$ $$v^{\prime }=\frac{v_1+v_2}{1+\frac{v_1{v}_2}{c^2}}$$ $$v^{\prime }=\frac{3/4+3/4}{1+9/16}$$ $$v^{\prime }=\frac{6/4}{25/16}$$ $$v^{\prime }=\frac{6}{4}\cdot \frac{16}{25}$$ $$v^{\prime }=\frac{96}{100}$$ $$v^{\prime }=\frac{24}{25}>\frac{9}{10}$$

Therefore the dart will catch up.

Question 3. Relativistic Lagrangian

$$L(x,\dot{x})=-m{c}^2\sqrt{1-\frac{{\dot{x}}^2}{c^2}}-\frac{1}{2}k{x}^2$$

Equations of motion

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x}=0$$ $$-m{c}^2{\left(1-\frac{{\dot{x}}^2}{c^2}\right)}^{1/2}-\frac{1}{2}k{x}^2$$ $$\frac{\partial L}{\partial \dot{x}}=\frac{2m{c}^2\dot{x}}{2c^2}{\left(1-\frac{{\dot{x}}^2}{c^2}\right)}^{-1/2}$$ $$\frac{\partial L}{\partial \dot{x}}=\frac{m\dot{x}}{\sqrt{1-\frac{{\dot{x}}^2}{c^2}}}$$ $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}{\left(1-\frac{{\dot{x}}^2}{c^2}\right)}^{-1/2}$$ $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=m\ddot{x}{\left(1-\frac{{\dot{x}}^2}{c^2}\right)}^{-1/2}-\frac{{\ddot{x}}^2}{2c^2}m\dot{x}{\left(1-\frac{{\dot{x}}^2}{c^2}\right)}^{-3/2}$$ $$\frac{\partial L}{\partial x}=-\frac{1}{2}2kx$$ $$\frac{\partial L}{\partial x}=-kx$$

Euler Lagrange Equation

$$m\ddot{x}{\left(1-\frac{{\dot{x}}^2}{c^2}\right)}^{-1/2}-\frac{{\ddot{x}}^2}{2c^2}m\dot{x}{\left(1-\frac{{\dot{x}}^2}{c^2}\right)}^{-3/2}+kx=0$$

At non-relativistic speeds

$$m\ddot{x}{\left(1-\cancel{\frac{{\dot{x}}^2}{c^2}}\right)}^{-1/2}-\cancel{\frac{{\ddot{x}}^2}{2c^2}m\dot{x}{\left(1-\frac{{\dot{x}}^2}{c^2}\right)}^{-3/2}}+kx=0$$ $$m\ddot{x}+kx=0$$

Question 4. Lagrangian Mechanics

a. Find the Lagrangian of the system +y is up, +x is to the right

$$L=T-U$$ $$T=\frac{1}{2}m{v}^2,U=mgh$$ $$x=L\cos \theta ;\dot{x}=\dot{L}\cos \theta -L\dot{\theta }\sin \theta$$ $$y=L\sin \theta ;\dot{y}=\dot{L}\sin \theta +L\dot{\theta }\cos \theta$$ $$T=\frac{1}{2}m({\sqrt{(\dot{L}\cos \theta -L\dot{\theta }\sin \theta {)}^2+(\dot{L}\sin \theta +L\dot{\theta }\cos \theta {)}^2}}^2$$ $$T=\frac{1}{2}m((\dot{L}\cos \theta -L\dot{\theta }\sin \theta {)}^2+(\dot{L}\sin \theta +L\dot{\theta }\cos \theta {)}^2)$$ $$T=\frac{1}{2}m(({\dot{L}}^2{\cos }^2\theta +L^2{\dot{\theta }}^2{\sin }^2\theta -2\dot{L}L\dot{\theta }\cos \theta \sin \theta )\dots$$ $$+({\dot{L}}^2{\sin }^2\theta +L^2{\dot{\theta }}^2{\cos }^2\theta +2\dot{L}L\dot{\theta }\sin \theta \cos \theta ))$$ $$T=\frac{1}{2}m(({\dot{L}}^2{\cos }^2\theta +L^2{\dot{\theta }}^2{\sin }^2\theta )+({\dot{L}}^2{\sin }^2\theta +L^2{\dot{\theta }}^2{\cos }^2\theta )$$ T=\frac{1}{2}m(\dot{L}^{2}(\cos ^{2}\theta + \sin ^{2}\theta) + L\dot{\theta}^{2}(\cos ^{2}\theta + \sin ^{2}\theta)) $$ $$ T = \frac{1}{2}m(\dot{L}^{2}+L\dot{\theta}^{2}) $$U=mg(L\cos \theta )$$ $$L=\frac{1}{2}m({\dot{L}}^2+L{\dot{\theta }}^2)+mg(L\cos \theta )$$

Sub in $\alpha$ and apply small angle approx.

$$L=\frac{1}{2}m({\alpha }^2+L{\dot{\theta }}^2)+mgL$$

Solve the Euler-Lagrange Equation

$$L=\frac{1}{2}m{\alpha }^2+\frac{1}{2}mL{\dot{\theta }}^2+mgL\cos \theta$$ $$\frac{\partial L}{\partial \dot{\theta }}=mL\dot{\theta }$$ $$\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}=m\dot{L}\dot{\theta }+mL\ddot{\theta }$$ $$\frac{\partial L}{\partial \theta }=mg\dot{L}\cos \theta -mgL\sin \theta$$