Convergence of integral and summation for Time taken for complete revolution around vertical circle

Question

I want to find Time taken to complete Vertical circular motion by Particle of mass $m$

So I proceed as follows

Applying work energy theorem

$$-mgR(1-\cos(\theta)=\frac12 mv^2-\frac12 mu^2$$

$$\implies v^2=u^2-2gR(1-\cos\theta)$$

$$\implies \omega=\frac{\sqrt{u^2-2gR(1-\cos\theta)}}{R}$$

$$\implies \frac {d\theta}{dt}=\frac{\sqrt{u^2-2gR(1-\cos\theta)}}{R}$$

$$\implies \frac {d\theta}{\sqrt{u^2-2gR(1-\cos\theta)}}=\frac{dt}{R}$$

Let T be the time taken for complete revolution

Now, Integrating both sides

$$\int_{0}^{2\pi}\frac {d\theta}{\sqrt{u^2-2gR(1-\cos\theta)}}=\int_{0}^{T}\frac{dt}{R}$$

$$\implies T=2R\int_{0}^{\pi}\frac{d\theta}{\sqrt{u^2-4gR\sin^2\theta}}$$

$$\implies T=\frac{2R}{u}\int_{0}^{\pi}\frac{d\theta}{\sqrt{1-\frac{4gR}{u^2}\sin^2\theta}}$$

On evaluating this elliptic integral we get,

$$T=\sum_{n=0}^{\infty} \left(\frac{2}{u}\right)^{2n+1}\left(\frac{(2n-1)!!}{2^n n!}\right)^2 (gR)^n$$

Here $T$ denotes Time taken for complete revolution. Block will complete full circle only if $u\geq \sqrt{5gR}$. So my question is $T$ will be Real only when $u\geq \sqrt{5gR}$ so This condition should exist in integral as well as summation for $T$ but I'm not able to see it. Also Wolfram alpha evaluates integral only if $u\geq \sqrt{5gR}$

Answer 1

Call problem $A$ the one you have drawn: the tracks may only provide a normal force radially inwards. Call problem $B$ the similar problem except that cart and track are replaced by bead on a wire. In $A$, the cart may fall off the track. In $B$, the 'cart' may not fall off the 'track'.

For problem $A$:

Minimum initial speed to reach height of $2R$: at least $\sqrt{4gR}$ (exact number not relevant to the arguments here)

Minimum initial speed to complete a loop: $\sqrt{5gR}$.

For problem $B$:

Minimum initial speed to reach height of $2R$: $\sqrt{4gR}$.

Minimum initial speed to complete a loop: $\sqrt{4gR}$.

Your equations of motion are those of problem $B$, since you have fixed $r=R$. The problems are identical when $u>\sqrt{5gR}$, when the normal force of the track is inwards over the whole loop. The problems are not identical if $u<\sqrt{5gR}$.

You get sensible answers (real and finite integral) for $\sqrt{4gR}<u<\sqrt{5gR}$, because in $B$, the cart does complete the loop. We know that in $A$, the cart does not for $u<\sqrt{5gR}$, so we have found sensible answers to a different problem.

Your elliptic integral has a singularity at $\sqrt{4gR}$. Wolfram agrees. This indicates $T \to \infty$ in problem $B$, as $u \to \sqrt{4gR}$: the bead takes infinitely long to reach the top of the loop.

Answer 2

Both the integral and the series converge if and only if $u > 2 \sqrt{g R}$, but your formula for $T$ is correct only if $u \geq \sqrt{5 g R}$. This discrepancy can be understood as follows:

The first condition only ensures that the particle has enough energy to reach the height $2R$ (it can be found by equating $2 R m g$ and $\frac{1}{2} m u^2$).

But your formula for $T$ can only be correct if the particle slides along the loop at all times. This only happens if the sum of the centrifugal force and the radial component of the gravitational force is always non-negative, i.e. $m R \omega^2 + m g \cos (\theta) \geq 0$. Using the conservation of energy to replace $\omega$ we can rewrite this inequality as $\frac{u^2}{g R} \geq 2 - 3 \cos(\theta)$. For $\theta = \pi$ the right-hand side is maximal, so we obtain the second (stronger) condition $\frac{u^2}{g R} \geq 5 ~ \Leftrightarrow~ u \geq \sqrt{5gR}$.

For $2 \sqrt{gR} < u < \sqrt{5gR}$ the particle has enough energy to reach the top of the loop, but it will detach itself from the track before that happens. Since your entire calculation is based on the energy conservation law (which contains less information than the original equations of motion), the stronger condition for the applicability of your result never appears and must be found separately.