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`.

Answer

`(14F)/(11m)`

Explanation

The forces, acceleration angular acceleration of disc and rod are shown in figure `a` and `b` .
The equation for disc `F'xxr=3/2mr^(2)alpha`……….i
`F'-f=ma_(2)`……ii
`a_(2)=ralpha_(2)`………iii

The equations or bar, `P-P'=ma` .......iv
`(F+F')L/2=(ML^(2))/12 alpha_(1)` ..........v
As point `A` on the bar and disc are same, we have
`a_(2)=ralpha_(2)=a-L/2 alpha_(1)` .............vi
On solving eqn i to vi we get `a=(8F)/(11m),alpha_(1)=(24F)/(22mL)`
`|a_(A)|=a_(2)=a-L/2alpha_(1)=(2F)/(11m)`
`|a_(B)|=a+L/2alpha_(1)=(14F)/(11m)`