Time taken to slide down a circle (friction)

[eddiezhang]

It’s been a blind manipulation of symbols to try and solve the ODE that was derived using forces. If anyone has a direct route to solve non linear ODE then it should be shared too.

Furthermore what do the rules say about AI in the report? PF and whatever math forum he’s visiting are effectively that as far as the OP is concerned. We are just a bit more reserved in our response.

AI is technically allowed, but the point of the report is to show a) understanding of the math used and b) solving things from the ground up (which is why I was discouraged from using energy concepts).

 

[eddiezhang]

Thanks. I like the symmetry of the abbreviation which truly conveys the symmetry of the idea it's trying to express.

I thought I might gain some insight on how this has already been approached and searched for the swing time of a pendulum starting from the horizontal which is this problem in the no-friction limit. The first thing that popped up was, as is usually the case with searches nowadays, the AI Overview which I show below.

View attachment 355723
Clearly, here "AI" stands for Artificial Ignorance since an amplitude of 90° is outside the realm of the "small angle" approximation in which T = 2π * sqrt(L/g) can be used.

After digging a bit deeper, look what I found in the PF archives. Follow the link given in post #5 to the Wikipedia article, understand what is involved in getting the answer to the frictionless case, toss in the friction terms and solve. Easier said than done.

Thanks, I will investigate this.

Was this problem given to you by a mathematician?

I have to write this for a math class. This was the scenario I chose to investigate, but if it doesn't resolve itself nicely or is too much effort, I can always just change topics.

 

[PeroK]

AI is technically allowed, but the point of the report is to show a) understanding of the math used and b) solving things from the ground up (which is why I was discouraged from using energy concepts).

Energy is fundamental. Once you go beyond Newtonian mechanics, there are no forces and you need energy-based methods: Lagrangian, Hamiltomian etc.

 

[kuruman]

I have to write this for a math class. This was the scenario I chose to investigate, but if it doesn't resolve itself nicely or is too much effort, I can always just change topics.

I see. I would be curious to know whether the person in charge of your class has actually solved this problem analytically before presenting to you as a choice and, if so, how.

 

[haruspex]

At the least, we can use dimensional analysis to discover the form of the relationship (assuming the start and finish angles are fixed): $t=\sqrt{\frac rg}f(\mu)$. Likewise the velocity at the bottom, but I feel the most interesting result is the angle at which it will come to rest (having started at the horizontal), since that will be a function of $\mu$ only.
So in principle we could use simulation with arbitrary values of r and g (both=1, say) to build up a table of results.
Doing that, I found the critical value for just making it to the bottom is very close to $\mu=0.6$.

 

[kuruman]

At the least, we can use dimensional analysis to discover the form of the relationship (assuming the start and finish angles are fixed): $t=\sqrt{\frac rg}f(\mu)$. Likewise the velocity at the bottom, but I feel the most interesting result is the angle at which it will come to rest (having started at the horizontal), since that will be a function of $\mu$ only.
So in principle we could use simulation with arbitrary values of r and g (both=1, say) to build up a table of results.
Doing that, I found the critical value for just making it to the bottom is very close to $\mu=0.6$.

I used the "bottom" expression $$\omega(\theta)= \left \{\left[ \frac{v_0^2}{R^2}- \frac{2g}{R}\frac{\mu}{(4\mu^2+1)}\right]e^{-\mu \pi} + \frac{2g}{R} \frac{(2\mu^2\cos\theta -\mu\sin\theta+\cos\theta)e^{2\mu \theta}}{4\mu^2+1} \right\}^{1/2}e^{-\mu~\theta}\tag{1}$$ derived here. Equation (1) gives the angular speed of the sliding block as a function of $\theta$ starting from the horizontal position $\left(\theta =-\frac{\pi}{2}\right)$ with initial speed $v_0.$ Setting $v_0=0$ and $\theta=0$, the angular speed at the bottom of the track for this case is $$\omega(0)= \left \{\left[- \frac{2g}{R}\frac{\mu}{(4\mu^2+1)}\right]e^{-\mu \pi}+ \frac{2g}{R} \frac{(2\mu^2 +1)}{4\mu^2+1} \right\}^{1/2}.\tag{2}$$ To find at what value of $\mu$ this is equal to zero, we set the right hand side of equation (2) equal to zero and see if we can solve the resulting equation for $\mu.$ After the obvious simplifications, we get $$\begin{align}
& 2\mu^2 +1=\mu e^{-\mu \pi} \nonumber \\
& \ln(2\mu^2 +1)=\ln(\mu) -\mu \pi. \nonumber \\
\nonumber \end{align}$$The left-hand side in the last equation is always positive whilst the right-hand side is always negative. It appears that the block cannot come to rest at the bottom even if it starts from rest at the top.

Shown below is $\omega(\theta)$ for several values of $\mu$ with $g=10~\text{m/s}^2$ and $R=1~\text{m}.$

If there is an error in the derivation of equation (1), I have not been able to find it.