A bob of mass m is suspended by a light string of length L It is imparted a horizontal - Physics - Units and Measurements

Question

A bob of mass m is suspended by a light string of length L It is
imparted a horizontal velocity vo 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
Obtain an expression for (i) vo; (ii) the speeds at points B and C;
(iii) the ratio of the kinetic energies (KB/KC) at B and C Comment
on the nature of the trajectory of the bob after it reaches the
point C

Decoder_2 · Accepted Answer

Dear student,
let the velocity at the topmost point be u m/sas the particle is  in vertical circular motion:so the centripetal acceleration is being provided by mg alone as string has slackedso mg = mu2Lso u = gL at the top most point (when string slacks)
speed at Cnow applying the energy conservation betweenA and C       we have mv022+0 = mu22 + mg(2L)so mv022 = 2mgL2 + mg(2L)so v0=  5gL
speed at Afor point Capplying conservation ofenergymv022+0 = muB22 + mg(L)so uB = 3gL ( velocity at B)b) KEB KEC = mvB22mu22 = v2Bu2 = 3gLgL  =31The object will complete the circular path and its trajectory will be circular as string has slacked at the highest point
Regards