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A mass attached to one end of a string crosses top-most point on a vertical circle with critical  speed, its centripetal acceleration when string becomes horizontal  will be (where, g= gravitational acceleration)

Answer:

Explanation:

 we know that,

 velocity of particle at top -most point on vertical circle,

 $(V_{top})= \sqrt{3rg}$    ...........(i)

 But centripetal acceleration (a_c)=$\frac{V^{2}}{r}$

 $\therefore$    $a_{c}=\frac{(\sqrt{3rg})^{2}}{r}$      [from Eq. (i)]

 $a_{c}=\frac{3rg}{r}$

 a_c  = 3g