Subfield: Mechanics


Question: A frictionless track contains a loop of radius $R=0.5 \mathrm{~m}$. Situated on top of the track lies a small ball of mass $m=2 \mathrm{~kg}$ at a height $h$. It is then dropped and collides with another ball of mass $M=5 \mathrm{~kg}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_e1013618489f33954a52g-1.jpg?height=361&width=797&top_left_y=1725&top_left_x=661)

The coefficient of restitution for this collision is given as $e=\frac{1}{2}$. Now consider a different alternative. Now let the circular loop have a uniform coefficient of friction $\mu=0.6$, while the rest of the path is still frictionless. Assume that the balls can once again collide with a restitution coefficient of $e=\frac{1}{2}$. Considering the balls to be point masses, find the minimum value of $h$ such that the ball of mass $M$ would be able to move all the way around the loop. Both balls can be considered as point masses.

Solution:
Let the angle formed by $M$ at any moment of time be angle $\theta$ with the negative y-axis. The normal force experienced by $M$ is just
$$
N=M g \cos \theta+M \frac{v(\theta)^{2}}{R}
$$

by balancing the radial forces at this moment. Now, applying the work energy theorem, we have

$$
\begin{gathered}
\int-\mu\left[M g \cos \theta+M \frac{v(\theta)^{2}}{R}\right] R \mathrm{~d} \theta=\frac{1}{2} M v(\theta)^{2}-\frac{1}{2} M v_{0}^{2}+M g R(1-\cos \theta) \\
\Rightarrow-\mu\left[M g \cos \theta+M \frac{v(\theta)^{2}}{R}\right] R=\frac{M}{2} \frac{\mathrm{d}\left(v(\theta)^{2}\right)}{\mathrm{d} \theta}+M g R \sin \theta
\end{gathered}
$$

Rearranging, we have

$$
\frac{\left.\mathrm{d}\left(v(\theta)^{2}\right)\right)}{\mathrm{d} \theta}+2 \mu v(\theta)^{2}=-2 g R(\sin \theta+\mu \cos \theta)
$$

Let $v^{2}(\theta)=y$. Thus we have a first order linear ODE of the form

$$
\frac{\mathrm{d} y}{\mathrm{~d} \theta}+P(\theta) y=Q(\theta)
$$

This is easily solvable using the integrating factor $e^{\int P(\theta) \mathrm{d} \theta}$. Here the integrating factor is

$$
e^{\int 2 \mu \mathrm{d} \theta}=e^{2 \mu \theta}
$$

So multiplying by the integrating factor, we get

$$
\begin{gathered}
\int \mathrm{d}\left(e^{2 \mu \theta} y\right)=\int-2 g R(\sin \theta+\mu \cos \theta) e^{2 \mu \theta} \mathrm{d} \theta \\
\Rightarrow y=\frac{\int-2 g R(\sin \theta+\mu \cos \theta) e^{2 \mu \theta} \mathrm{d} \theta}{e^{2 \mu \theta}}
\end{gathered}
$$

Now we use the well known integrals

$$
\begin{aligned}
& \int e^{a x} \sin x \mathrm{~d} x=\frac{e^{a x}}{1+a^{2}}(a \sin x-\cos x) \\
& \int e^{a x} \cos x \mathrm{~d} x=\frac{e^{a x}}{1+a^{2}}(a \cos x+\sin x)
\end{aligned}
$$

(These integrals can be computed using integration by parts.) Thus, plugging and chugging these integration formulas into our expression for $y$ and integrating from $\theta=0$ to $\theta=\phi$, we have upon solving

$$
v^{2}(\phi)-v_{0}^{2}=\frac{-2 g R}{1+4 \mu^{2}}\left[\left(3 \mu \sin \phi+\left(2 \mu^{2}-1\right) \cos \phi-\left(2 \mu^{2}-1\right) e^{-2 \mu \phi}\right]\right.
$$

where $v_{0}$ is the velocity at $\phi=0$. Solving gives us the velocity as a function of angle covered

$$
v(\phi)=\sqrt{v_{0}^{2}-\frac{2 g R}{1+4 \mu^{2}}\left[\left(3 \mu \sin \phi+\left(2 \mu^{2}-1\right) \cos \phi-\left(2 \mu^{2}-1\right) e^{-2 \mu \phi}\right]\right.}
$$

But to cover a complete circle, at the top most point

$$
N=m g-\frac{m v^{2}(\pi)}{R} \geq 0 \Rightarrow v(\pi) \leq \sqrt{g R}
$$

Thus

$$
v_{0} \leq \sqrt{g R\left[1+\frac{2\left(1-2 \mu^{2}\right)}{1+4 \mu^{2}}\left(1+e^{-2 \mu \pi}\right)\right]}
$$



From the previous expression,

$$
v_{0}=\frac{m(1+e) \sqrt{2 g h}}{M+m} \geq \sqrt{g R\left[1+\frac{2\left(1-2 \mu^{2}\right)}{1+4 \mu^{2}}\left(1+e^{-2 \mu \pi}\right)\right]}
$$

Hence

$$
h \geq \frac{R(M+m)^{2}}{2 m^{2}(1+e)^{2}}\left[1+\frac{2\left(1-2 \mu^{2}\right)}{1+4 \mu^{2}}\left(1+e^{-2 \mu \pi}\right)\right]
$$

We get $h \geq 72.902 \mathrm{~m}$ and we are done.

Final answer: $72.902$ m