The vertical circle is the most important topic in circular motion for JEE. At any angle theta from the bottom:

Radial equation: T - mgcos(theta) = $mv^2$/L (at angles where T points toward centre and the component of mg away from centre is mgcos(theta))

Energy conservation: $v^2$ = $v_{bottom}^2$ - 2gL(1 - cos(theta))

At the bottom (theta = 0): T = $mv^2$/L + mg (maximum tension) At the top (theta = 180): T = $mv^2$/L - mg (minimum tension) Tension difference: $T_{bottom}$ - $T_{top}$ = 6mg (universal, regardless of speed)

String vs Rod:

• String: T >= 0 at all points. Critical condition at top: v >= sqrt(gL). This gives $v_{bottom}$ >= sqrt(5gL).
• Rod: Can push (T can be negative). $v_{top}$ can be 0. This gives $v_{bottom}$ >= sqrt(4gL) = 2*sqrt(gL).

If $v_{bottom}$ < sqrt(5gL) (string): The ball leaves the circular path at the angle where T = 0, which is always above the centre line. After leaving, it follows projectile motion.