A uniform slender bar `AB` of mass `m` is suspended as shown from a small cart of the same mass `m`. Neglecting the effect of the friction, determine the accelerations of points `A` and `B` immediately after a horizontal force `F` has been applied at `B`.

Explanation

Let the acceleration of bar and cart be a and a', respectively: also `alpha` be the angular acceleration of the bar.

Writing newton's equation of motion
For the bar: `F-F'=ma`……i
For the cart `F'=ma'`..........ii
Torque equation for rod about centre of mass
For the rotation of the rod `(F+F')L/2=(ML^(2))/ alpha`
`F+F'=(MLalpha)/6`..........iii
The acceleration of point `A` on the rod will be same as the acceleration of the cart
`veca_(A)=veca_(A,C)=veca_(C)`
`-a'=alphaL/2-a` or `a=a -L/2 alpha`......iv
on solving eqn i, ii, iii, and iv we get
`a=(7F)/(5m)` and `alpha(18F)/(5mL)`
`|a'|=|a_(A)|=a-L/2alpha=(2F)/(5m)`
`|a_(B)|=a+L/2alpha=(16F)/(5m)`