Two particles of mass m each are attached to a light rod of length d, one at its centre and the other at a free end, The rod is fixed at the other end and is rotated in a plane at an angular speed `omega`. Calculate the angular momentum of the particle at the end with respect to the particle at the centre

Answer

(i) `sqrt(2ag)/(3)` (ii) `(3v)/(g)` (iii) `(2v)/(g)`

Explanation

`2mg - T = 2m.a` ….(i)
`T - mg = m.a` ….(ii)
On solving eq. (i) & (ii) `a = (g)/(3)`
(a) `v_(B) = sqrt(u^(2) + 2as) = sqrt(0 + 2((g)/(3))a) = sqrt((2ag)/(3))`
(b) `s = ut + (1)/(2) t^(2), a = 0 + (1)/(2)((g)/(3)) t^(2), t = sqrt((6a)/(g)) = (3v)/(g)`
(c ) `t = (2v)/(g)`