Find the acceleration of rod $A$ and wedge $B$ if the ratio of the mass of wedge to that of the rod equals $\eta$ , and the friction between the contact surfaces are negligible.

 

Let in any time the rod displaces down by $y$ , the corresponding horizontal displacement of wedge will be $x$ . We have

                          $y=x \tan \theta$ .         …(i)

Differentiating both sides of equation (i), we get

                          $\frac{dy}{dt} = \frac{dx}{dt}\tan \theta$

or                      $v_{rod} = v_{wedge} \tan \theta$       …(ii)

Also                   $a_{rod} = a_{wedge} \tan \theta$      …(iii)

Let mass of the rod is $m$ , then mass of wedge, $M = \eta m$ .

By Newton’s second law :

For vertical motion of rod;

                 $mg-N \cos \theta = m{a}_{rod}$          …(iv)

For horizontal motion of wedge;

                $N \sin \theta = M{a}_{wedge}$

                             $=M\left(\frac{a_{rod}}{\tan \theta }\right)$

Solving equations (iv) and (v), we get

                               $a_{rod} = \frac{g}{\left[1+{\displaystyle \frac{M}{m}}co{t}^2\theta \right]}$

                                       $=\frac{g}{\left[1+{\displaystyle \eta }co{t}^2\theta \right]}$

and                         $a_{wedge} =\frac{a_{rod}}{\tan \theta }= \frac{g}{\left[\tan \theta +{\displaystyle \eta }co{t}^2\theta \right]}$ .