# Useful Derivation for Labs Involving Rolling Balls Down an Inclined Plane

• MHB
• Ackbach
In summary, the conversation discusses the problem of large experimental errors in beginning mechanics physics labs and how it can be caused by factors such as friction and comparing apples to oranges. The theory of balls rolling down inclines is also discussed, with the solution being derived using conservation of energy and rotational kinetic energy. A simplified version of the solution is also presented, which involves differentiation instead of integration. The idea of using a general moment of inertia and finding that only the shape of the object is significant in rolling under gravity is also mentioned.
Ackbach
Gold Member
MHB
Problem:
In beginning mechanics physics labs, it is very often the case that you get large experimental error. This can be due to a number of factors, friction, of course, being a major player. However, it can also be the case that you might be comparing apples to oranges. For example, if you're rolling balls down inclines, what should you expect to see for the position function? If you do frame-by-frame video analysis with rulers, and come up with an acceleration quite different from your theory, you're going to chalk that up as merely experimental error. But it can be helpful to compare apples to apples. So what is the theory of balls rolling down inclines?

A ball of solid sphere shape rolls without slipping down a straight inclined plane that makes an angle of $\theta$ with the horizontal. It starts from rest at the point $x_{0}$, where $x$ is measured positively up the plane. Find the position $x=x(t)$.

We use conservation of energy, including rotational kinetic energy, to obtain the differential equation of motion. Let $v$ be the speed of the ball. Let $y$ be the height of the ball at time $t$, with $h$ being the original height of the ball. Let $m$ be the mass of the ball, and let $r$ be its radius. Then Conservation of Energy tells us that
$$\frac{1}{2}\,mv^{2}+mgy+\frac{1}{2}\,I\omega^{2}=mgh.$$
The no-slip condition tells us that $v=\omega r$. The moment of inertia for a solid sphere of radius $r$ is
$$I=\frac{2}{5}\,mr^{2}.$$
Hence, the energy conservation law becomes
\begin{align*}
\frac{1}{2}\,mv^{2}+mgx \sin( \theta)+ \frac{1}{2} \left( \frac{2}{5}\,mr^{2} \right) \left( \frac{v}{r} \right)^{2}&=mgh\\
\frac{1}{2}\,mv^{2}+mgx \sin( \theta)+\frac{1}{5} \,mv^{2}&=mgh\\
\frac{7}{10} \,v^{2}+gx \sin( \theta)&=gh\\
v^{2}&= \frac{10g}{7}[h-x \sin( \theta)].
\end{align*}
Let $a=\sqrt{10g/7}$, so that we obtain
$$v=a\sqrt{h-x\sin(\theta)},$$
or
$$\frac{dx}{\sqrt{h-x\sin(\theta)}}=a\,dt.$$
We can solve this separable ODE simply by integrating both sides as follows (changing $x$ to $\xi$ and $t$ to $\tau$ for the dummy variables of integration):
$$\int_{x_{0}}^{x}\frac{d\xi}{\sqrt{h-\xi\sin(\theta)}}=a\int_{0}^{t}d\tau.$$
A simple $u$-substitution conquers the LHS integral, and the RHS integral is very straight-forward: it comes out to $at$. Hence, we have that
$$\frac{2\sqrt{h-x_{0}\sin(\theta)}-2\sqrt{h-x\sin(\theta)}}{\sin(\theta)}=at.$$
Note that $h=x_{0}\sin(\theta)$, and thus this simplifies immediately down to
$$-\frac{2\sqrt{h-x\sin(\theta)}}{\sin(\theta)}=at.$$
Multiplying by $\sin(\theta)$ yields $-2\sqrt{h-x\sin(\theta)}=a\sin(\theta)\,t.$ Squaring both sides yields
$$4[h-x\sin(\theta)]=a^{2}\sin^{2}(\theta)\,t^{2},$$
or
$$4h-a^{2}\sin^{2}(\theta)\,t^{2}=4\sin(\theta)\,x.$$
Dividing through by $4\sin(\theta)$ yields
$$x=\frac{h}{\sin(\theta)}-\frac{a^{2}}{4}\,\sin(\theta)\,t^{2}.$$
Now $h=x_{0}\sin(\theta)$, and hence $h/\sin(\theta)=x_{0}$. Hence, we have
$$x=x_{0}-\frac{a^{2}}{4}\,\sin(\theta)\,t^{2}.$$
Plugging back in for $a$ yields
$$x=x_{0}-\frac{10g}{4\cdot 7}\,\sin(\theta)\,t^{2}=x_{0}-\frac{5g}{14}\,\sin(\theta)\,t^{2}.$$
Hence, the answer to the problem is
$$\boxed{x(t)=x_{0}-\frac{5g}{14}\,\sin(\theta)\,t^{2}}.$$
The well-known kinematic equation if you ignore rotational kinetic energy would yield
$$x(t)=x_{0}-\frac{g}{2}\,\sin(\theta)\,t^{2}.$$
So you can see that the $t^{2}$ term is off by $2/14$ or $1/7$.

Conclusion:
Experiments using a camera to video balls rolling down inclined planes yield a typical $2\%$ error with the more accurate version, compared to over a $100\%$ error after $1$ second of rolling using the less sophisticated number. It definitely pays to have a more accurate theory in mind when you are analyzing your experimental data.

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Thanks @Ackbach, what physics forum can we move this to?

Greg Bernhardt said:
Thanks @Ackbach, what physics forum can we move this to?
STEM Educators and Teaching, I think.

Greg Bernhardt
The derivation in post #1 could be streamlined a bit. As already shown, one gets the expression for the speed squared using energy conservation, $$v^{2}= \frac{10g}{7}[h-x \sin( \theta)].$$The acceleration in the ##x## direction (along the incline) is $$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac{1}{2}\frac{d}{dx}\left(v^2\right)=\frac{1}{2} \frac{10g}{7}[- \sin( \theta)]=-\frac{5g}{7}\sin(\theta).$$Assuming that ##v_0=0## and since the acceleration is constant, it follows that $$x(t)=x_0-\frac{5g}{14}\sin(\theta)~t^2.$$

pbuk, PeroK and Ackbach
kuruman said:
The derivation in post #1 could be streamlined a bit. As already shown, one gets the expression for the speed squared using energy conservation, $$v^{2}= \frac{10g}{7}[h-x \sin( \theta)].$$The acceleration in the ##x## direction (along the incline) is $$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac{1}{2}\frac{d}{dx}\left(v^2\right)=\frac{1}{2} \frac{10g}{7}[- \sin( \theta)]=-\frac{5g}{7}\sin(\theta).$$Assuming that ##v_0=0## and since the acceleration is constant, it follows that $$x(t)=x_0-\frac{5g}{14}\sin(\theta)~t^2.$$
True, no doubt. There might be a very few students who could do that trick, but I doubt there would be many. And as R. L. Moore would have said, let the student use any valid method they choose.

Ackbach said:
True, no doubt. There might be a very few students who could do that trick, but I doubt there would be many. And as R. L. Moore would have said, let the student use any valid method they choose.
I wholeheartedly agree. It is a simple "trick" that involves differentiation and avoids integration when the acceleration is constant. I routinely showed it to my students as an alternative and let them choose whatever method they felt comfortable with. Not exactly R. L. Moore's teaching style.

Ackbach
I've never taught maths or physics, but I like the idea of using a general moment of inertia of the form ##I = kmr^2##. Then finding that only ##k## is significant. I.e. rolling under gravity depends only on the shape of the object, not on its size or mass.

vela and kuruman
PeroK said:
##\dots~## rolling under gravity depends only on the shape of the object, not on its size or mass.

Ackbach said:
There might be a very few students who could do that trick, but I doubt there would be many.
I disagree. Differentiation using the chain rule is much simpler than integrating using a substitution - particularly if you are distracted by introducing the variable ## a ## to mean something other than acceleration - why would you choose that? Also it is all much easier if you ignore the "starting height" and just consider the difference in height which we can write immediately as ## (\sin \theta) x ##.

We then have from conservation of energy:
$$\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = (mg\sin\theta)x$$
Substituting ## I = \frac{2}{5} m r^2 ## and ## \omega^2 = \frac{v^2}{r^2} ##:
\begin{align} \frac{1}{2} m v^2 + \frac{1}{2} \frac{2}{5} m r^2 \frac{v^2}{r^2} &= (mg\sin\theta)x \\ \frac{5}{10} v^2 + \frac{2}{10} v^2 &= (g\sin\theta)x \\ \frac{7}{10} v^2 &= (g\sin\theta)x \\ \end{align}
Differentiating wrt ## t ## using the chain rule:
$$\frac{7}{5} v \dot v = (g\sin\theta) \dot x$$
Noting ## v = \dot x ## and acceleration ## a = \dot v ##:
\begin{align} a &= \frac{5}{7} g\sin\theta \\ x &= \frac{1}{2} a t^2 = \frac{5 g\sin\theta}{14} t^2 \end{align}

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vanhees71, malawi_glenn and berkeman
pbuk said:
I disagree. Differentiation using the chain rule is much simpler than integrating using a substitution - particularly if you are distracted by introducing the variable ## a ## to mean something other than acceleration - why would you choose that? Also it is all much easier if you ignore the "starting height" and just consider the difference in height which we can write immediately as ## (\sin \theta) x ##.

We then have from conservation of energy:
$$\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = (mg\sin\theta)x$$
Substituting ## I = \frac{2}{5} m r^2 ## and ## \omega^2 = \frac{v^2}{r^2} ##:
\begin{align} \frac{1}{2} m v^2 + \frac{1}{2} \frac{2}{5} m r^2 \frac{v^2}{r^2} &= (mg\sin\theta)x \\ \frac{5}{10} v^2 + \frac{2}{10} v^2 &= (g\sin\theta)x \\ \frac{7}{10} v^2 &= (g\sin\theta)x \\ \end{align}
Differentiating wrt ## t ## using the chain rule:
$$\frac{7}{5} v \dot v = (g\sin\theta) \dot x$$
Noting ## v = \dot x ## and acceleration ## a = \dot v ##:
\begin{align} a &= \frac{5}{7} g\sin\theta \\ x &= \frac{1}{2} a t^2 = \frac{5 g\sin\theta}{14} t^2 \end{align}
I guess we'll have to agree to disagree. I am looking at this from the teacher perspective, as well as from the perspective of someone who was never the "prodigy" who "just got it". I had to work really hard to understand anything. As a result, I understand students when they don't get something right away.

Many tricks can be shown, and perhaps the student might remember it later. But to expect any student to come up with this derivation on their own, never having seen this kind of trick before, would be well-nigh impossible. Not one in a thousand students would figure this out on their own.

Ackbach said:
It starts from rest at the point $x_{0}$, where $x$ is measured positively down the plane.
Ackbach said:
Hence, the answer to the problem is
$$\boxed{x(t)=x_{0}-\frac{5g}{14}\,\sin(\theta)\,t^{2}}.$$
So the ball rolls up the plane?

Ackbach
vela said:
So the ball rolls up the plane?
Good catch. I've obviously got a sign error somewhere...

vela said:
So the ball rolls up the plane?
Much easier to change coordinate systems than to re-do all the math.

How is the chain rule of differentiation a "trick?

pbuk said:
Differentiation using the chain rule is much simpler than integrating using a substitution
That is exactly how I teach my students how to do this. They learned the chain rule of differentiation the year before when they were 16-17y old. No idea in what part of the world students learn fancy integrals before chain rule of differentiation. Not in my country at least.

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vanhees71, pbuk and kuruman
Ackbach said:
Many tricks can be shown, and perhaps the student might remember it later. But to expect any student to come up with this derivation on their own, never having seen this kind of trick before, would be well-nigh impossible. Not one in a thousand students would figure this out on their own.
Who's expecting students to come up with this "trick" on their own? The idea behind STEM teaching is to give students a toolbox with a variety of methods and approaches to problem-solving. Testing at the intro level is about seeing whether students have mastered the use of these tools and are able to use them correctly. It is not about sitting back and seeing who can construct tools on their own to perform the task. We know what they cannot do, let's ascertain what they can do.

As students progress through the intermediate and graduate level of courses, the tools in the toolbox become more sophisticated. Eventually, students are indeed expected to fashion their own tools through a process called a "Ph.D. dissertation." That's my naive, perhaps, view of STEM education.

kuruman said:
Who's expecting students to come up with this "trick" on their own? The idea behind STEM teaching is to give students a toolbox with a variety of methods and approaches to problem-solving. Testing at the intro level is about seeing whether students have mastered the use of these tools and are able to use them correctly. It is not about sitting back and seeing who can construct tools on their own to perform the task. We know what they cannot do, let's ascertain what they can do.

As students progress through the intermediate and graduate level of courses, the tools in the toolbox become more sophisticated. Eventually, students are indeed expected to fashion their own tools through a process called a "Ph.D. dissertation." That's my naive, perhaps, view of STEM education.
Well, sure. Any trick can become part of the curriculum somewhere. Not every trick is taught everywhere. I, for one, was never taught to solve any ODE the way pbuk does in post #9. I was taught other, equivalent methods. I do not view my education as lacking simply because I wasn't taught that trick.

In the end, the only integrals I used in the original post involved either integrating a constant, or using ##u## substitution. Neither is very difficult, and both should be taught in integral calculus somewhere. So I don't see that the method I presented here is worse than pbuk's method. It's not better, either. Just different.

malawi_glenn said:
How is the chain rule of differentiation a "trick?That is exactly how I teach my students how to do this. They learned the chain rule of differentiation the year before when they were 16-17y old. No idea in what part of the world students learn fancy integrals before chain rule of differentiation. Not in my country at least.
There actually are calculus books (Apostol comes to mind) that focus on integration before differentiation. I never learned from them.

But pbuk is actually not just using the chain rule in post #9: it's the chain rule in the middle of a derivation to set up an integral. pbuk still has to integrate (as anyone would eventually have to do somewhere in solving this problem), which is what happens in post #9 going from line (4) to line (5). So post #9 is definitely more than just the chain rule, though it certainly does use that.

The integration I use is hardly fancy: a ##u## substitution in combination with the power rule to handle the square root, in addition to the integral of a constant. These should be taught in any integral calculus course.

Ackbach said:
pbuk still has to integrate (as anyone would eventually have to do somewhere in solving this problem), which is what happens in post #9 going from line (4) to line (5)
No I don't, having shown that acceleration is constant ## (a = \frac{5}{7} g\sin\theta) ## I simply substitute it into the SUVAT equation ## s = \frac{1}{2} a t^2 ## - the integration is already done for me.

malawi_glenn, PeroK and hutchphd
malawi_glenn said:
How is the chain rule of differentiation a "trick?

That is exactly how I teach my students how to do this. They learned the chain rule of differentiation the year before when they were 16-17y old.
Yes, that's exactly when I learned it. And when I was 17-18 years old I was taught to use it in exactly this method for solving cylinder/disc/ball rolling down plane problems, which used to come up at least every other year in the Mechanics paper of either physics or further maths.

malawi_glenn
You can avoid integration and differentiation altogether starting with pbuk's (3)
$$\frac 7{10} v^2 = (g \sin\theta) x \quad \Rightarrow \quad v^2 = 2\left(\frac 57 g\sin\theta\right)x$$ and comparing this result to the kinematic equation
$$v^2 = v_0^2 + 2a\Delta x.$$

PeroK and malawi_glenn
Ackbach said:
for one, was never taught to solve any ODE the way pbuk does in post #9. I was taught other, equivalent methods
Is integrating a constant twice a trick? And as mentioned, as soon as you have a constant acceleration, one can use the SUVAT equations which are taught in the first 2-3 weeks of basic physics.

I find it very hard to believe that "your" integral with u-substitution is anywhere taught before integrating a constant twice.

Ackbach said:
Not one in a thousand students would figure this out on their own.
Isn't that the reason why we have schools in the first place?

pbuk said:
And when I was 17-18 years old I was taught to use it in exactly this method for solving cylinder/disc/ball rolling down plane problems, which used to come up at least every other year in the Mechanics paper of either physics or further maths.
I basically do this.
## v^2 = kx ## where ##k## is a constant that depends on the physical properties of the solid body and the geometry of the problem.
Then I ask "what will we get if we differentiate both sides with respect to time ##t##"?
After 5 or so minutes, almost everybody will arrive at ##2v \dfrac{\mathrm{d} v}{\mathrm{d} t} = k \dfrac{\mathrm{d}x}{\mathrm{d} t} ## well perhaps someone will forget about that factor of 2.
Then I ask them do think about the relationships between the rate of change of velocity and the rate of change of position, what are those called in physics?
After that, almost everybody will arrive at ##2va = kv## and the conclusion that ##2a = k##.
In this way, they are discovering this method (yes it is a method, not a trick) mostly on their own, with some guidance from me and each other.
Since I started doing it this way, the success rate is much higher on exams regarding these problems.

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pbuk
Ackbach said:
The integration I use is hardly fancy: a u substitution in combination with the power rule to handle the square root, in addition to the integral of a constant. These should be taught in any integral calculus course.
I guess your approach to solving ODEs is different from mine. If I see $$\ddot x+\text{(const)}=0,$$ I do not integrate, but immediately write down the solution as $$x(t)=x(0)+\dot x(0)t-\frac{1}{2}\text{(const)}t^2.$$ I was taught that, once I solve an ODE and get the general solution, I do not have to solve it again and again, but write down the most general solution and apply the initial conditions.

In this particular problem, the acceleration is constant. For a displacement ##\Delta x## of the mass, the constant acceleration is $$\ddot x=\frac{v^2}{2\Delta x}$$as required by the kinematic equations. Once one obtains ##v^2## from energy conservation, the rest follows. The chain rule is not really needed when the acceleration is constant. When the acceleration is not constant, one can always write $$\ddot x=\frac{1}{2}\frac{d}{dx}\left(v^2\right)$$

I favor simplicity, but I respect your agreeing to disagree.

vela said:
You can avoid integration and differentiation altogether starting with pbuk's (3)
$$\frac 7{10} v^2 = (g \sin\theta) x \quad \Rightarrow \quad v^2 = 2\left(\frac 57 g\sin\theta\right)x$$ and comparing this result to the kinematic equation
$$v^2 = v_0^2 + 2a\Delta x.$$
And in general if ##I = kmr^2## we have$$v^2 = 2\frac{g\sin \theta}{k+1}x$$And$$a = \frac{g\sin \theta}{k+1}$$

malawi_glenn
pbuk said:
No I don't, having shown that acceleration is constant ## (a = \frac{5}{7} g\sin\theta) ## I simply substitute it into the SUVAT equation ## s = \frac{1}{2} a t^2 ## - the integration is already done for me.
Look, this is really not worth quibbling about. If you want to claim that you don't actually have to compute an antiderivative like $$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\quad\text{for } n\not=-1,$$ fine. But in effect you must integrate one way or the other, because that is how you solve the necessary differential equations.

For constant acceleration from rest the distance traveled is the area of a triangle of height ##at## and base length ##t##. That can be calculated without integration.

malawi_glenn said:
Is integrating a constant twice a trick? And as mentioned, as soon as you have a constant acceleration, one can use the SUVAT equations which are taught in the first 2-3 weeks of basic physics.

I find it very hard to believe that "your" integral with u-substitution is anywhere taught before integrating a constant twice.Isn't that the reason why we have schools in the first place?I basically do this.
## v^2 = kx ## where ##k## is a constant that depends on the physical properties of the solid body and the geometry of the problem.
Then I ask "what will we get if we differentiate both sides with respect to time ##t##"?
After 5 or so minutes, almost everybody will arrive at ##2v \dfrac{\mathrm{d} v}{\mathrm{d} t} = k \dfrac{\mathrm{d}x}{\mathrm{d} t} ## well perhaps someone will forget about that factor of 2.
Then I ask them do think about the relationships between the rate of change of velocity and the rate of change of position, what are those called in physics?
After that, almost everybody will arrive at ##2va = kv## and the conclusion that ##2a = k##.
In this way, they are discovering this method (yes it is a method, not a trick) mostly on their own, with some guidance from me and each other.
Since I started doing it this way, the success rate is much higher on exams regarding these problems.
Integrating a constant twice is not what I would call a "trick". I would call that a standard theorem that calculus students must learn.

From the context, I'm guessing that the "SUVAT" equations are the kinematic equations for constant acceleration? I've never seen that abbreviation before. You need to keep in mind that different people in different parts of the world have very different education systems, different traditions, etc.

I don't know what your point is in talking about whether ##u## substitution is taught before or after integrating a constant twice. To solve the problem the way I did you need both.

We certainly do have schools to teach us basic methods. Tricks are fine and dandy, but they have a way of being taught less methodically. All you need to do is read Feynman's book Surely You're Joking, Mr. Feynman to see that differentiation under the integral sign is not a universally taught "trick". Some people see it in their education, others don't.

Different students learn things differently. I'm not really talking about Gardner's silliness of multiple intelligences (mostly debunked, as I understand, but still taught in many "education" departments). I just mean that an approach to learning a concept might work very well for one student, and totally flop for another. I would find your method of teaching the physics here somewhat confusing, but no doubt many other students would learn from it very well.

PeroK said:
For constant acceleration from rest the distance traveled is the area of a triangle of height ##at## and base length ##t##. That can be calculated without integration.
True, but it doesn't generalize very well. You could think of the problem geometrically as you've done here, or in a calculus manner. The calculus approach of solving a second-order linear ODE is much more generalizable.

PeroK
The main point of the original post was that in a lab situation, rolling balls down inclines is a great lab. But you need to have the right theory or you're going to get too much experimental error - too much difference between theory and experiment. The main point of the OP was not that such-and-such a method was the latest and greatest way to solve the problem. There are, no doubt, hundreds of valid ways to solve the problem, each with trade-offs in ease-of-calculation, ease-of-understanding, generalizability, and probably other factors. To me, those are incidental. I just wanted to solve the problem somehow, to provide a justification for the final expression of ##x(t)##. I would say mission accomplished.

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Ackbach said:
True, but it doesn't generalize very well. You could think of the problem geometrically as you've done here, or in a calculus manner. The calculus approach of solving a second-order linear ODE is much more generalizable.
It seems a bit hard on high school students I'd they can't calculate the area of a triangle until they've studied integral calculus: or, solve constant acceleration problems until they've studied ODES!

PeroK said:
It seems a bit hard on high school students I'd they can't calculate the area of a triangle until they've studied integral calculus: or, solve constant acceleration problems until they've studied ODES!
Well, that has nothing to do with what I was trying to say, and is what I would call a strawman.

If you're in algebra-based physics, you learn algebra-based methods. If you're in calculus-based physics, you can use calculus-based methods. I'm certainly not advocating forcing students to use methods they haven't learned, yet. You have to build the edifice, one brick at a time, and you can't put the roof on if the foundation isn't laid.

Ackbach said:
I just wanted to solve the problem somehow
Ackbach said:
I would say mission accomplished.
The title of the thread is "Useful derivation". And this what we are arguing about here. And what didactical advantages this integration method is compared to differentiation method + SUVAT.

Imo, it is more like "Somewhat cumbersome derivation", because the differentiation method is quicker and that its easier to differentiate than integrate.

vanhees71
malawi_glenn said:
The title of the thread is "Useful derivation". And this what we are arguing about here. And what didactical advantages this integration method is compared to differentiation method + SUVAT.

Imo, it is more like "Somewhat cumbersome derivation", because the differentiation method is quicker and that its easier to differentiate than integrate.
The derivation is useful in the lab. That is, it is useful for the derivation to have been accomplished so that, in the lab, you get far less experimental error. The method of deriving the result is basically not the point of the OP at all. I derive it so that the student has confidence in the result, but that's all. The point of the OP is that having the correct formula reduces your experimental error.

Greg Bernhardt and vanhees71
I'd also say for 1D problems the most simple way to derive the equation of motion is to use energy conservation and derive the corresponding equation once by time. In your case you have
$$\frac{m}{2} \dot{x}^2 + \frac{I}{2 R^2} \dot{x}^2 - m g x \sin \theta= \frac{M_{\text{eff}}}{2} \dot{x}^2 - m g x \sin \theta=E=\text{const}, \quad M_{\text{eff}}=m+\frac{I}{R^2}=\frac{7}{5}m.$$
Taking the time derivative gives
$$\dot{x} (M_{\text{eff}} \ddot{x}-m g \sin \theta)=0.$$
Now you can have either ##x=\text{const}##, but it's obvious that this is not a solution, or
$$M_{\text{eff}} \ddot{x}=m g \sin \theta \; \Rightarrow \; x(t)=x_0 + v_0 t +\frac{m g}{2M_{\text{eff}}} t^2.$$

malawi_glenn
I have re-framed the OP to focus on what was the main event, in my mind. The main event is the lab, and trying to reduce experimental error. My apologies if that wasn't clear. I'm not interested in continuing any discussions about my method of solving the theoretical problem. I would argue it's a valid way, and that's what's needed.

malawi_glenn and Greg Bernhardt
Well, but also in a lab the theoretical foundation should be made as clear as possible, and an overly complicated derivation for sure doesn't help the understanding of the physics. When I was a student, we had to prepare the theory underlying the experiment in the lab beforehand, and for mechanics labs I always used Hamilton's principle to prepare. Once the "colloquium" checking whether we were prepared took only 5 minutes, because instead of being a complicated derivation with the Lagrangian it were 3 lines. A good theoretical methodology always helps also the experiments!

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