A light rigid rod of length `4 m` is connected rigidly with two identical particles each of mass `m = 2 kg`. the free end of the rod is smoothly pivoted at `O`. The rod is released from rest from its horizontal position at `t = 0`. Find the

angular acceleration of the rod at `t = 0` (in `rad s^(-2)).`

Answer

3

Explanation

Torque equation. The internal forces and the reaction forces at `O`. Then the net torque about `O` is
`tau_(0)=mgl+mg(2l)=3mgl`
Which produces an angular acceleraton `alpha`. Using Newton's 2nd law of rotation `tau=Ialpha`
we have `3mgl=I_(0)alpha` where `I_(0)=ml^(2)+m(2I)^(2)=5ml^(2)`
This gives `alpha=(3g)/(5l)=(3xx10)/(5x2)=3rads^(-1)`
Force equation: Refering `FBD`. we have `(mg+mg)-rR=ma_(1)+ma_(2)`

Substitutig `a_(1)=Ialpha` and `a_(2)=2/Ialpha`, we have
`R=m(2g-3Ialpha)`
`=2(2xx10^(-3)xx2x3)=2(20-18)=4N`