In a vertical circle

Circular Motion - OCR A-Level Physics

In a vertical circleCircular motion in a vertical plane where the net centripetal force varies with position due to gravity., speed varies because gravity does work on the object. The speed is greatest at the bottom and least at the top.
At the top of a vertical loop: weight and any contact/tension force both act downward (towards centre). Minimum condition for maintaining contact: mg = mv^$2_{top}/$r, giving $v_{top} =$ $\sqrt{gr}.$
At the bottom of a vertical loop: normal/tension force acts upward, weight acts downward. Net upward $force = N - mg = mv^2_{bottom}/r$.
Using energyThe capacity to do work. Measured in joules (J). conservation between top and bottom: $v^2_{bottom} = v^2_{top} + 4gr.$
If speed at the top falls $below \sqrt{gr}$, the object loses contact with the track (or the string goes slack).

Worked Example

A ball on a string of length 0.80 m moves in a vertical circle. What is the minimum speed at the top for the string to remain taut?

Show Solution

At the top, the minimum condition is when $tension = 0$, so weight provides all the centripetal forceThe resultant force directed towards the centre of a circular path that causes an object to move in a circle. It is not a separate force but the net force providing circular motion..

$mg =\;\text{m}v^2/r.$

$v^2 = gr = 9.81 \times 0.80 = 7.85.$

$v = \sqrt{7.85} = 2.8 m s^{-1}.$

Answer

v_{min} = 2.8 $m s^{-1}$ at the top

Common Mistake MEDIUM

Wrong: At the top of a vertical circle, setting the centripetal force equal to the tension or normal force alone.
Right: At the top, BOTH weight and tension/normal force act towards the centre: mg + $T = mv^2/r. The minimum speed occurs$ when T = 0, giving $mg = mv^2/r$.