First slide

Vertical circular motion

Question

A small sphere of mass m is tied to one end of a string of length l. The other end of the string is attached to a fixed point o . The sphere is released from position A when the string is horizontal and the sphere starts moving in a vertical circle. When the string makes an angle of θ with horizontal, centripetal acceleration of the sphere is,

Easy

A

2g sin θ
B

g cos θ
C

g sin θ
D

2gcos θ

Solution

If V be the speed of the sphere at position B,
Then conserving energy, $\frac{1}{2} {mv}^{2} = mglsinθ$
$\Rightarrow \frac{v^{2}}{l} = 2 gsinθ$
$∴$ Centripetal acceleration at B,
$a_{n} = \frac{v^{2}}{l} = 2 gsinθ$

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