A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v_(o) at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for The speeds at points B and C.

A bob of mass m is suspended by a light string of length L. It is impa

1.	A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v_(o) at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for The speeds at points B and C.
Answer» Solution :It is clear from Eq. (6.14) `v_(c )= sqrt(gL)` At B, the energy is `E=(1)/(2)mv_(B)^(2)+mgL` Equatting this to the energy at A and employing the RESULT from (i), namely `v_(0)^(2)= 5gL`, `(1)/(2)mv_(B)^(2)+mgL= (1)/(2)mv_(0)^(2)` `(5)/(2)mgL` `v_(B)= sqrt(3gL)`.