Question

A uniform bar AB of mass m and a ball of the same mass are released from rest from the same horizontal position. The bar hinged at end A. There is gravity downwards. What is the distance of the point from point B that has the same acceleration as that of ball, immediately after release. ?

Answer 1

Moment of inertia rod = $\frac{1}{3}m{\ell }^2$

Torque of the rod hinged of end $A$ in $mg\frac{\ell }{2}$

$Z=I\alpha$

$mg\frac{1}{2}=\frac{1}{3}m{\ell }^2\alpha$

$\alpha =\frac{3}{2\ell }g$

$a=\alpha x$

$g=\alpha x$

$a=\frac{g}{\alpha }=\frac{g}{\frac{3}{2\ell }g}=\frac{2\ell }{3}$ from point $A$

From point $B=\ell -\frac{2\ell }{3}=\frac{\ell }{3}$