Theoretical Mechanics: Fall-2025
HW 3: Due W 10/8 Day 11
- Constant Acceleration by Separation of Variables
Calculate: Treat Newton's 2nd law as a separable differential equation and solve for the velocity and position as a function of time of an object that is all of the following:
- moving in one dimension,
- not initially at the origin of coordinates,
- moving with a non-zero initial speed,
- experiences a constant force.
Reflect: Do your answers look familiar? If yes, from where? If not, how would you have to modify these equations to be similar to equations you know?
- Separable ODE linear + constant
Solve the differential equation:
\(\frac{dv}{dt}=-b-cv\) where \(v(t=0)=v_0\)
- Distance Formula in Curvilinear Coordinates
The distance \(\left\vert\vec r -\vec r\,{}'\right\vert\) between the point \(\vec r\) and the point \(\vec r'\) is a coordinate-independent, physical and geometric quantity. But, in practice, you will need to know how to express this quantity in different coordinate systems.
Find the distance \(\left\vert\vec r -\vec r\,{}'\right\vert\) between the point \(\vec r\) and the point \(\vec r'\) in rectangular coordinates.
Show that this same distance written in cylindrical coordinates is: \begin{equation*} \left|\vec r -\vec r\,{}'\right| =\sqrt{s^2+s\,{}'^2-2ss\,{}'\cos(\phi-\phi\,{}') +(z-z\,{}')^2} \end{equation*}
Hint: You may want to use the textbook: GMM: Change of Coordinates
Show that this same distance written in spherical coordinates is: \begin{equation*} \left\vert\vec r -\vec r\,{}'\right\vert =\sqrt{r'^2+r\,{}^2-2rr\,{}' \left[\sin\theta\sin\theta\,{}'\cos(\phi-\phi\,{}') +\cos\theta\cos\theta\,{}'\right]} \end{equation*}
Hint: You may want to use the textbook: GMM: Change of Coordinates
- Now assume that \(\vec r\,{}'\) and \(\vec r\) are in the \(x\)-\(y\) plane. Simplify the previous two formulas.