Physics · Laws Of Motion · NEET
A common mistake is thinking the object can slow to a stop at the top. But at the top the object is still moving in a circle, so it still needs a centripetal (inward, downward) force. The only downward forces available are weight mg and string tension T, so mg + T = mv²/R. A real string can only pull, so T cannot be negative; the smallest possible case is T = 0. Setting T = 0 gives mg = mv_top²/R, so v_top = √(gR). If the speed were less, the required centripetal force would be smaller than mg, and gravity would pull the object inward off the circular path — it would fall like a projectile before reaching the top.
Start from the top condition v_top² = gR. Now use energy conservation between the top and bottom of the loop. The vertical height difference is the diameter, 2R. So ½m·v_bottom² = ½m·v_top² + mg(2R). Substitute v_top² = gR: v_bottom² = gR + 4gR = 5gR, giving v_bottom = √(5gR). This is why √(5gR) is the minimum launch speed at the bottom — anything less and the object cannot keep the string taut all the way to the top.
The string goes slack (T = 0) before the object reaches the top. Once the string is slack, no tension acts, gravity is the only force, and the object leaves the circular track and becomes a projectile. It does NOT complete the loop. For a string or a bucket of water this is the failure case — the water spills. This is different from a rigid rod, which can push as well as pull.
No — and NEET tests this difference. A string or a chain can only pull (tension ≥ 0), so you need v_top = √(gR) and v_bottom = √(5gR). A light rigid rod (or the inside of a smooth tube/track) can also push inward or outward, so it can support the object even at zero speed at the top. For a rod, the minimum speed at the top is 0, and the minimum speed at the bottom is √(4gR) = 2√(gR). Always check whether the problem says 'string/thread/chain' (√5gR) or 'rod/tube' (2√gR).
At the bottom the centripetal direction is upward, so tension must both support the weight and provide the centripetal force: T_bottom = mg + mv_bottom²/R. At the top the centripetal direction is downward, so gravity helps and tension only needs to make up the rest: T_top = mv_top²/R − mg. The bottom also has the largest speed. Both effects make T_bottom the maximum, so the wire is most likely to snap at the lowest point. Useful result: T_bottom − T_top = 6mg for any vertical circle.
What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?
A body, initially at rest, slides down a frictionless track from a height h and just completes a vertical circle of diameter AB = D. The height h is equal to:
A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For a string, chain, or bucket, the minimum speed at the lowest point is v = √(5gR) and at the highest point it is v = √(gR), where R is the radius and g is the acceleration due to gravity.
A string can only pull, so tension cannot be negative. The smallest allowed value is zero. At T = 0 the entire centripetal force at the top comes from gravity: mg = mv²/R, which gives the minimum top speed √(gR).
√(gR) for a string. The speed² at the bottom (5gR) is exactly 5 times the speed² at the top (gR).
A rigid rod or tube can push as well as pull, so the object can have zero speed at the top. The minimum bottom speed for a rod is √(4gR) = 2√(gR), not √(5gR).
At the lowest point, where T = mg + mv²/R and the speed is greatest. The tension difference between bottom and top is always 6mg.
