A body crosses the topmost point of a vertical circle with critical speed. What will be its centripetal acceleration when the string is horizontal.

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Solution

The correct option is C
3g

Step1: Given data
The object at the top of the vertical circle has a speed equal to critical speed.
Step2: Formula used
The critical speed at the top of a vertical circle, $v_{c} = \sqrt{g l} [w h e r e g = a c c e l e r a t i o n d u e t o g r a v i t y, l = r a d i u s o f t h e v e r t i c a l c i r c l e]$
At a point in a horizontal circle, when the string becomes horizontal, the speed of the object,
$v^{2} = u^{2} + 2 g h [w h e r e v = f i n a l s p e e d, u = i n i t i a l s p e e d, g = a c c e l e a r t i o n d u e t o g r a v i t y, h = c h a n g e i n h e i g h t]$
Step3: Calculating the final speed.
According to the given data, the speed of the object at the topmost point $= \sqrt{g l}$
This will act as the initial speed for when the string becomes horizontal i.e. $u = \sqrt{g l}$
Therefore the speed of the object at this point will be
$v^{2} = {(\sqrt{g l})}^{2} + 2 g h$
The change in height of the object from the initial point is equal to,
$v^{2} = g l + 2 g l = 3 g l$
Step4: Calculating centripetal acceleration
$a_{c} = \frac{v^{2}}{l} a_{c} = \frac{3 g l}{l} a_{c} = 3 g$
The centripetal acceleration when the string becomes horizontal is 3g.
Hence, option C is the correct answer.

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