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**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 $x=x(t)$.

**Answer:**

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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