Vertical Circular MotionJEE Advanced

Loop, oscillate, or fly off the circle — interactive Physics simulation for IIT-JEE.

Concept

A bob on a string moving in a vertical circle trades kinetic energy for height. The string can only pull, so above the horizontal the motion survives only while $v^2 \ge gr\cos\alpha$ . Depending on the launch speed the bob completes the loop, oscillates like a pendulum, or leaves the circle and becomes a projectile.

Key formula

v^2(\theta) = u^2 - 2gr(1-\cos\theta), \qquad u_{min} = \sqrt{5gr}, \qquad T - mg\cos\theta = \frac{mv^2}{r}

Derivation

Energy conservation from the bottom gives $v^2(\theta)$ . At the top, the minimum condition is $T=0$ : gravity alone supplies the centripetal force, so $v_{top}^2 = gr$ .

Energy back to the bottom: $u_{min}^2 = v_{top}^2 + 4gr = 5gr$ .

Three regimes: $u^2 \le 2gr$ → speed dies below the horizontal (oscillates); $2gr < u^2 < 5gr$ → string slackens above the horizontal (leaves); $u^2 \ge 5gr$ → completes.

Scenarios to explore

Vertical Circular Motion — Complete the loop, oscillate, or fly off — √(5gr) decides.

Real-world applications

Roller-coaster loops (rail can push, so v_top can be less than √(gr)).
Buckets of water swung overhead.
Pendulum release problems and trapeze mechanics.

JEE exam tips

T_bottom − T_top = 6mg for a full loop — independent of speed! A classic one-liner.
After leaving the circle, the bob is a projectile launched at angle α above horizontal.
For the rod/tube version replace 5gr → 4gr everywhere.

Common mistakes

Using √(5gr) for a bead on a rigid rod — rods can push, so u_min = √(4gr) (v_top = 0 suffices).
Forgetting tension at the bottom is mg + mu²/r, not mg.
Applying the leave-circle analysis below the horizontal (string can't slacken there).

Exam traps to avoid

√(5gr) is the speed at the BOTTOM; the speed at the top is √(gr).
The bob leaves the circle where T = 0, not where v = 0 — the two coincide only at exactly u = √(5gr).