The actual period of a pendulum at 90°. Looking for the correct formula

Question

Do you have access to any scientific experiment which gives the period of a pendulum when the angle is $90^\circ$:

this article says $T$ varies to about $18\%$ up to $90^\circ,$ so for a seconds pendulum the complete period would be roughly $2.36.$

1. I d'like to know the ratio at $90^\circ$ with a good approximation, and the span it can vary in relation to air friction, if the bob is not massive.

2. Also, I'd like to know the max. speed of the bob at $P_0$ and how it compares with the speed of same ball sliding without friction on a circular surface, would it be the same?

What I am trying to understand here is rather complex, I'll try to explain:

If the block is placed on an air-track or other frictionless surface, the block would acquire a certain speed, momentum and KE in the opposite direction, right? same amount of momentum is lost by the bob and we can find speed loss.

3. Isn't the same amount of speed lost by the bob when it is attached to a pendulum?

4. Is this an important factor or may be the most important factor (more relevant than air friction or friction at the pivot) for the increase of the period when the angle is greater than $30^\circ$?

5. I figured out that this factor can account for about one third of the $18\%$ increase, can you give a more accurate value?

What I am suggesting is that this (loss of horizontal momentum, which is made clear when the bob is not attached to a rod but slides on a block) is the most important factor that makes the actual period increase when the starting point deviates on the vertical direction.

Answer 1

REVISED ANSWER

I think you are trying to solve the problem of a small sphere of mass $m$ sliding down the circular face of a wedge of mass $M$. There is no friction between the sphere and the wedge and the wedge is free to move on a horizontal air track.

This problem is similar to block slides on smooth triangular wedge kept on smooth floor.Find velocity of wedge when block reaches bottom except that the flat slant face of the wedge is now the quadrant of a circle. When the sphere reaches the bottom of the slope it is travelling horizontally to the left with velocity $v$ while the wedge is moving to the right with velocity $V$. Energy and momentum are conserved so :
$$mv = MV$$ $$mv^2+MV^2=2mgR$$ therefore
$$v^2=\frac{2gMR}{m+M}$$ $$V^2=\frac{2gmR}{m+M}$$ where $R$ is the radius of the circle traced out by the centre of the small sphere - which is not the same as the radius $R'$ of the circular face of the block : $R=R'-r$ where $r$ is the radius of the sphere.

Note that if $M \gg m$ then $v^2 \approx 2gR$, which is the same as for a pendulum of length $R$. In this case the time of descent will also be the same as the quarter-period of a pendulum of length $R$ and amplitude $\theta_0=\frac{\pi}{2}$. Using the value for $T$ given in this webpage the time of descent is $1.8541\sqrt{\frac{R}{g}}$.

If $m \gg M$ then the bob drops approx. vertically down and the time of descent is $1.4142\sqrt{\frac{R}{g}}$.

If $m$ is comparable with $M$ then the time of descent will lie between the extremes of 1.4142 and 1.8541 in units of $\sqrt{\frac{R}{g}}$. The exact value will be very difficult to calculate - much harder than for the simple pendulum, the exact solution of which involves an elliptic integral for 'large' amplitudes.

ORIGINAL ANSWER

The wikipedia article for the Simple Pendulum gives formulas for the period of the pendulum, in terms of elliptic integrals or as power series, as a function of amplitude. It also gives a very fast-converging method using the arithmetic-geometric mean solution for the elliptic integral.

The speed of the bob at the lowest point is easily calculated using conservation of energy : $v^2=2gh$. Yes, this is the same as for a ball sliding (not rolling) down the inside of a spherical shell.

Your other questions (below your 2nd image) about a block on an air track is not at all clear. How does that relate to the pendulum?