If you are limiting the takeoff speed to prevent it bottoming out, then I suggest you lower the ramp. $45^\circ$ gives optimal range for a given takeoff speed (ignoring friction), but only if you don't care what the vertical component of the velocity is on landing.
At $30\: \mathrm{mph}$, with a $45^\circ$ jump, you say it doesn't bottom out. The vertical component of the velocity on landing has approximately the same magnitude as on take off (air resistance losses), which would be $30 \; \cos(45^\circ)$ = about $26\: \mathrm{mph}$.
To maximise the range, then, we need to keep this vertical component ($v_y = 26\: \mathrm{mph}$), whilst letting the overall speed ($v$) increase to $45\: \mathrm{mph}$.
$v^2 = v_x^2 + v_y^2$
$v_x^2 = v^2 - v_y^2 = 45^2 - 26^2 = 1349$
$v_x = \sqrt{1349} = 36.7 \mathrm{mph} $
So we can calculate the angle from the $x$ and $y$ components of the velocity:
$\tan(\theta) = v_y / v_x = 26 / 36.7 = 0.71$
$\theta = \arctan(\theta) = \arctan(0.71) = 35^{\circ}$
So based on that information, try $35^\circ$, with a speed at the ramp of $45\: \mathrm{mph}$.
Actually you proved, with the third equation, that the translational speed does not depend on the radius $r$ of
the cylinder, or on its mass $m$, but only on $g$ and $\alpha$. Indeed it reads like a differential equation defining the motion of the cylinder.
If the translational speed of the cylinder axis does not depend on the radius, then the angular speed has to depend on it, since they are related by the relation $v=\omega r$.
I do not see how that fact can be overcome by any computation
including the total momentum of the system. This momentum can be
computed from the speed at distance $l$, but will depend on the radius
$r$, because it is squared in the moment of inertia, but not in the
angular speed. Getting the total momentum by other means, such as resolving the differential equation will not do any better. Imho (but I have not practiced this sport for a long time).
But having $r$ in the expression of the angular speed is not a crime.
BTW What are you supposed to do with the coefficient of friction $\mu$ ?
Do you have to check the cylinder does not slip?
Best Answer
I gather that the large source of error you are worried about is the ability of the experimenter to accurately hit the start/stop button on the stopwatch at the start/stop of the ball's journey down the ramp.
What is the approximate magnitude of error we'd expect?
Before I directly answer your question, let's estimate how bad the experimental error will be (reasonably close, without actually doing the experiment).
Typical human reaction time is 2-3 tenths of a second, somewhat increasing with age. I would expect the error to be smaller at the top of the ramp (easier to synchronize with "3..2..1..go!"). I would also expect the experimenter to be able to somewhat anticipate when the ball will reach the end of the ramp (because they can see the ball rolling towards it), and these timings may even somewhat cancel each other out. So for argument's sake (not to be used as an actual estimate; that's what science is for), let's be optimistic and say the total amount of reaction error would be around 100ms. For a typical length of ramp used in such an experiment, ball bearing travel times would be on the order of 1 second (1000ms).
Thus, your reaction error is:
$\frac{val_{exp} - val_{known}}{val_{known}} = \frac{(1000+100)ms - 1000ms}{1000ms} = 10\%$
Note: When applied to a calculation of g, the above error will actually be worse, on the order of $g \pm 17\%$, left as an exercise to the reader to verify.
So, yes, a 10% error is far from your desired 1%. Can we do better?
Use a longer ramp or a shallower incline
You'd think you could just use a longer ramp or a shallower incline. Unfortunately, to get < 1% error with a 100ms (total) error, you need a 10 second roll, the length of ramp required with a shallow ($10^{\circ}$) incline is given by:
$s = v_{i}t + \frac{1}2at^2$ but since $v_{i} = 0$,
$s = \frac{1}2at^2$
And we know $t = 10 sec$ and $a = g * \sin \theta = 9.81m/s^2 * \sin 10^{\circ} = 1.703m/s^2$
$s = 85.15 m$
I imagine that ramp length would have several practical limitations.
Can we do better?
Unfortunately, not much. Given the constraints in the question regarding "ordinary" measurements with rulers and stopwatches, a typical classroom experiment of this type will have a fairly large error. Using many trials and some statistical analysis (if this is a senior high school or university-level experiment) to get a confidence interval for your experimental value might improve the analysis, but not the actual error. Put another way, no math is capable of going back in time and pushing that stopwatch button for the experimenter. Plus, you asked how to improve the experiment, not the analysis.
However, as a teaching tool, that error is actually a rather valuable part of the experiment. It's one of those rare experiments where the value you're trying to measure (g) is very well known, and the sources of error are at the same time significant, but also easy to imagine, and not too difficult to estimate.
Remember, science isn't just about results. It's about questions. (Right, well, tell that to my grant application reviewers.)