Consider a skating ramp that is shaped as the bottom left quarter of a circle. A skater starts at the top of the ramp and skates down to the bottom. What is the normal force on the skater at each point in the trajectory? Ignore all considerations that would make this more realistic, such as resistive forces; it is a problem meant to think about elementary concepts like differential equations, conservation of energy, and circular motion.
Here is how I view the problem.
Let's measure the angle $\theta$ of the skater with respect to horizontal line passing through the center of the circle and the initial skater position.
There is a gravitational force on the skater $\overrightarrow{F_g}=-mg \hat{j}$. If we use another coordinate system then $\overrightarrow{F_g}=mg\sin(\theta)\hat{r} + mg\cos(\theta)\hat{\theta}$.
From the equations for circular motion:
$$\overrightarrow{a}=a_{R} \hat{r} + a_T \hat{\theta}=-r\theta'^2\hat{r}+r\theta''\hat{\theta}$$
Using Newton's second law in directions $\hat{r}$ and $\hat{\theta}$:
$$N-mg\sin(\theta)=ma_R=mr\theta'^2$$
$$mg\cos(\theta)=ma_T=mr\theta''$$
Note that $N,\theta,\theta',\theta''$ are all functions of $t$.
My questions are: is this the legitimate/correct approach to calculating the normal force?
If so, it looks like I have two differential equations. I know how to solve first and second order linear differential equations of various types. I am not sure if I know how to solve these two at this stage. I would just like to know what direction I should go in to learn to solve these equations, ie, is there a specific technique or do I need to know how to solve systems of equations, etc?
Note that using conservation of energy, I believe we can find the normal force as a function of height.
$$-\Delta U=\Delta K$$
$$-mg(y_f-y_i)=\frac{mv_f^{2}}{2}$$
$$y_i=R \implies v_T=v_f=\sqrt{2g(R-y_f)}$$
So we know the speed at each height, and we know it is the tangential speed because velocity is perpendicular to the circular motion. The result we obtained follows from the fact that work is only done by gravity, and so there is conservation of mechanic energy, and in this example this is independent of the shape of the ramp (since the normal force is always perpendicular to motion, and thus does zero work).
We know the radial acceleration is given by
$$a_R=-R \theta'^2$$
But we also know the relationship between angular speed and tangential speed:
$$v_T=\theta' R$$
$$\implies a_R=\frac{-2g(R-y_f)}{R}$$
At any point in the trajectory we can use Newton's second law:
$$mg\sin{\theta}-N=\frac{m(-2g(R-y_f)}{R}$$
$$\sin{\theta}=\frac{R-y_f}{R}$$
$$N=3mg\frac{R-y_f}{R}$$
This is the magnitude of the normal force as a function of height $y_f$. To find normal force as a function of time, we'd need to know how $y_f$ changes in time.
Best Answer
This problem is completely equivalent to the simple pendulum. Your ramp's normal force is then just the force along the rod or string of the pendulum. You can find the derivation of the equation of motion of the simple pendulum online. The problem is that this equation of motion is not analytically solvable, only the so called small-angle approximation is. But if the skater starts at $\theta = \pi/2$, this approximation is not valid and you have to resort to numerical solutions for the time evolution.