Mathematical method in school physics

Question

I have a simple physical task: there is a car, riding on an ellipse ($a=500$, $b=250$), with a constant absolute velocity ($|\vec{v}|$ constant). I need to find the maximum and minimum acceleration $\vec{a}$ in each point.

I represent the position vector $\vec{r}$ as a vector function $\vec{r}(t)$. Also, I know that velocity is the first derivative, $\vec{v}=\vec r\,'$, and acceleration is the second derivative, $\vec{a}=\vec r\,''$.

How should I approach this problem?

Answer 1

Acceleration is given by the formula $$ \vec{a}=\frac{dv}{dt}\vec{u}_T+\frac{v^2}{R}\vec{u}_{\perp} $$ where $\frac{dv}{dt}\vec{u}_T$ is the linear acceleration ($\vec u_T$ is the unit vector tangent to the trajectory) and $\frac{v^2}{R}\vec{u}_{\perp}$ the centripetal/normal acceleration ($\vec u_{\perp}$ is the unit vector perpendicular to $\vec u_T$). $R$ is the radius of curvature, that is the radius of the circumference that best approximates the car's trajectory in every point.

So, knowing that $v$ is constant, linear acceleration is always zero, so only the centripetal one counts, and it is greater when $R$ is smaller, which happens at the vertices of the major axis, so you have maximum acceleration at these two points. On the other hand you have minimum acceleration where $R$ is greater, that is at the vertices of the minor axis.

Answer 2

There is a nice discussion of the relevant computations at http://www2.math.umd.edu/~jcooper/math241/accel.pdf. The following is a quick summary of the relevant parts:

Suppose $r$ is a $C^2$ curve. Suppose $v(t) = \dot{r}(t) \neq 0$, and let $\tau(t) = { v(t) \over \|v(t)\|}$. Notice that $\|\tau(t)\| = 1$, and so $\langle \dot{\tau}(t), \tau(t) \rangle = 0$. Further, suppose $\dot{\tau}(t) \neq 0 $ and let $\nu(t) = { \dot{\tau}(t) \over \| \dot{\tau}(t) \| }$. Note that $v(t) = \|v(t)\| \tau(t)$, and so $a(t) = \dot{v}(t) = a_t(t) \tau(t) + a_n \nu(t) $, where $a_t(t) = \dot{s}(t)$, with $s(t) = \|v(t)\|$ and $a_n(t) = \|v(t)\| \| \dot{\tau}(t) \|$.

Now we need to parametrize the ellipse appropriately. If we let $p(\theta) = (a \cos \theta, b \sin \theta)$ we see that $p$ traces out the appropriate ellipse, but the speed is not constant. We need to find a curve $t \mapsto \theta(t)$ so that $ r(t) = p(\theta(t))$ satisfies $\|v(t)\| = \sigma $, where $\sigma$ is the constant speed.

We have $v(t) = (-a\sin \theta(t), b \cos \theta(t) ) \dot{\theta}(t)$, and so $\sqrt{(a \sin \theta(t))^2 + (b \cos \theta(t))^2}|\dot{\theta}(t)| = \sigma$. Hence if we let $\theta$ solve the differential equation $\dot{\theta}(t) = { \sigma \over \sqrt{(a \sin \theta(t))^2 + (b \cos \theta(t))^2} }$ subject to $\theta(0) = 0$, then $r$ traces out the ellipse with constant speed.

Since the speed is constant, we see that $a_t(t) = 0$. We have $\tau(t) = { (-a\sin \theta(t), b \cos \theta(t) ) \over \sqrt{(a \sin \theta(t))^2 + (b \cos \theta(t))^2} } $. Differentiating gives $\dot{\tau}(t) = { -( a b^2 \cos \theta(t), a^2 b \sin \theta(t)) \over ({(a \sin \theta(t))^2 + (b \cos \theta(t))^2})^{3\over 2} } {\sigma \over \sqrt{(a \sin \theta(t))^2 + (b \cos \theta(t))^2} } $.

Hence $\| \dot{\tau}(t) \| = {\sigma |ab| \over ({(a \sin \theta(t))^2 + (b \cos \theta(t))^2})^{3\over 2} }$, and so $\|a(t)\| = a_n(t) = {\sigma^2 |ab| \over ({(a \sin \theta(t))^2 + (b \cos \theta(t))^2})^{3\over 2} } $.