I believe the expression for motion in a circle, measured at the top and bottom of that circle,is $\frac{mv^2}{r} = T - mg$ where $mg$ is negative because it acts downwards at all times.
I am confused about the tension $T$ at positions on the circle that are not at the top or the bottom, especially at $\theta=$90 degrees to the vertical, for example.
Take a mass $m$ connected to a pivot by a rod or string and made to rotate in a vertical circle. At $\theta=$90 degrees the rod or string holding the mass would be horizontal.
At any positions on the circumference of the circle the horizontal component of the weight, $mg \cos\theta$, is equal to zero, and the vertical component is $mg\sin\theta=mg$
And does the centripetal force still act towards the centre, so the horizontal component must be $\frac{mv^2}{r}$?
In which case, at $\theta=90$ degrees are the weight and centripetal forces orthogonal?
Does this mean that if $\frac{mv^2}{r}$ is horizontal and is the resultant of the tension and $mg$, then in order to obtain a horizontal resultant there must be a horizontal component to the tension? Does this mean the tension must point upwards from the horizontal? Or is the tension actually horizontal?
If the latter is the case, does it mean that the tension vector always points towards the centre of the circle at all points on the circumference?