Energy In Vertical Circular Motion

Question

A small bead is whirled in a vertical circle of radius 2.5 m at the end of a string. What is the minimum speed it must have at the highest point so that the string remains just taut there?

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At the top of a vertical circle, the minimum-speed condition occurs when the string tension drops to zero and gravity alone provides the centripetal force needed for circular motion. Setting $mg = mv^2/r$ and cancelling mass gives the critical speed $v = \sqrt{gr}$. With $g = 10	ext{ m/s}^2$ and $r = 2.5	ext{ m}$, this is $v = \sqrt{10 	imes 2.5} = \sqrt{25} = 5	ext{ m/s}$. The option 25 m/s reports $v^2$ without taking the root. The option 2.5 m/s simply restates the radius and ignores $g$. The option 7 m/s loosely approximates $\sqrt{2gr}$, which is a different, non-critical condition. Below this critical speed the centripetal force required would exceed what gravity alone can supply, the string would slacken, and the bead would abandon its circular path and fall inward as a projectile. The very same condition governs the minimum speed needed to keep water in a bucket swung overhead, or for a vehicle to safely negotiate the top of a circular loop. This is the NCERT analysis of the critical speed at the top of a vertical loop. As a plausibility check, at this speed the required centripetal acceleration $v^2/r = 25/2.5 = 10	ext{ m/s}^2$ equals $g$, exactly the condition for the string tension to vanish.

Energy In Vertical Circular Motion

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