Here is a question that I thought of while doing a physics lab in class. We dropped a marble from a certain height down a ramp and around loop-the-loop and let it fly off of a table and then we were able to use the location at which it landed to work backwards to find it's initial velocity before going of of the ramp, then we could use energy to find the energy loss, then the force of friction, then the coefficient of friction.
If it helps we used an equation something like this:
$$E_{initial}=E_{final}$$
$$E_{gravitational}=E_{kinetic}+E_{rotational}-E_{friction}$$
$$mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2-F_{friction}∆d$$
Then we solved for the friction force. Now, I am only in highschool, so it's not a super advanced physics course and we assumed that the increase in normal force from the loop-the-loop (due to centripetal force) cancelled out with the decrease in the normal force from the ramp, so we just said that $F_n=mg$. But this is obviously not accurate. So here's my question:
How would you account for the constantly changing normal force around the loop?
I have though about this a bit, but it seems beyond me. This is only my first year taking calculus and here are my thoughts on how I could solve it:
If we could write an equation for the normal force as a function of time, then we could use the mean value theorem for integration to find the average normal force and I would think that that would work if we put it into the equation. The problem I found is getting to that point.
Could you please help me through this problem Thanks!
Best Answer
dmckee's comment is correct, this problem is hard (impossible?) to solve with pen and paper. But it's not too hard to set up, which hopefully will answer your question. There will advanced calculus by high school standards (including differential equations), no way around that, but I'll try to explain the equations at a high school level, I hope...
I'll skip the ballistic portion of the lab, which is fairly similar in setup but simultaneously more complicated (because the ball is not constrained to follow a particular path) and less complicated (because there is no rolling friction, only air drag). I'll include air drag, but you could get rid of this term if you wanted to simplify things (though at this point you'll be solving with a computer either way, so you may as well include it).
Start with conservation of energy. This means that at any point in time, the energy released from gravitational potential must be equal to the kinetic energy (translational & rotational) plus the work done (energy lost) so far by friction and drag.
$$U_{g}-U_{g,i} = K_t + K_r + W_f + W_d$$
We'll go through these term by term, but we need to set up a few things before that. First, what is our goal? Ultimately, we'll have two unknowns, a constant each for friction ($\mu$) and for drag ($C_d$, which I'll define later). There's also going to be a major intermediate result, which is to find $s(t)$: I'll call the distance travelled along the track $s$, so $s(t)$ is a function that describes how long it takes the ball to get to any given location on the track - clearly this is a useful quantity!
I'll point out that the speed of the ball is $\frac{{\rm d}s}{{\rm d}t}$. When we need to work with vectors, I'll use $\hat{x}$ & $\hat{y}$ coordinates, where gravity is in the $-\hat{y}$ direction. The other very important vector will be the normal vector of the track, $\hat{N}$. I'll write this vector as:
$$\hat{\bf N} = N_x(s)\hat{\bf x} + N_y(s)\hat{\bf y}$$
There is also the tangent vector, which we'll need at one point:
$$\hat{\bf T} = T_x(s)\hat{\bf x} + T_y(s)\hat{\bf y}$$
The two functions $N_x(s)$ and $N_y(s)$ depend on the shape of the track, but they are known: you can write them down using the known shape of the track and geometry.
Now we're ready to go:
So that's all the terms, now I'll just stuff them all into the first equation at the top of this answer (and put everything on one side)...
$$0 = mg\left(y(s) - y_0\right) + \frac{1}{2}m\left(\frac{{\rm d}s}{{\rm d}t}\right)^2 + \frac{1}{2}\frac{I}{r^2}\left(\frac{{\rm d}s}{{\rm d}t}\right)^2 +\\ \mu m\int_0^s\left[\frac{\left(\frac{{\rm d}s'}{{\rm d}t}\right)^2}{R(s')} - gN_y(s')\right]{\rm d}s' + \pi r^2 \rho C_d \int_0^s\left[\left(\frac{{\rm d}s'}{{\rm d}t}\right)^2\right]{\rm d}s'$$
This is quite a mess, but it is an integro-differential equation for $s(t)$. The idea is to find a function $s(t)$, when substituted into this equation, makes the right hand side equal to $0$. In this case, I'm almost certain that this cannot be done by hand (pen and paper) and instead you'd need to find an approximate solution using a computer. There are a couple of last complications before you can solve this. First, equations like this need what are called initial and boundary conditions. These give some extra constraints on the function $s(t)$. Here they're pretty straightforward, and just express that the length of the track is $L$, that the initial speed is $0$ and that the speed at launch time $t_L$ (which you would measure in the lab) is $v_L$ (which you would also measure in the lab). Here are the conditions:
Last but not least, there's the matter of the two unknowns we are trying to measure, $\mu$ and $C_d$. Unless my intuition is failing me, I think you'll end up with a system of equations with too many unknowns to solve, given your input. This is easy enough to remedy by making some additional timing and/or velocity measurements in the lab. Things are pretty dicey since $s(t)$ depends on $\mu$ and $C_d$, and you need to know the constants before finding $s(t)$ because you're finding an approximate solution. This ends up meaning you need to find solutions $s(t)$ for many combinations of $\mu$ and $C_d$ (probably in some clever way that helps find the right one quickly), then pick the one that works with your measurements.
So uh... I guess you can see why your teacher suggested a simple assumption to keep the problem from getting out of control!