Question

A uniform thin rod of mass M and length L is standing vertically along the y-axis on a smooth horizontal surface, with its lower end at the origin (0,0). A slight disturbance at t = 0 causes the lower end to slip on the smooth surface along the positive x-axis and the rod starts falling. The acceleration vector of centre of mass of the rod during its fall is[$\overline{R}$ is reaction from surface]

• A. ${\overline{a}}_{CM}=\frac{M\overline{g}+\overline{R}}{M}$
• B. ${\overline{a}}_{CM}=\frac{M\overline{g}-\overline{R}}{M}$
• C. $\overline{a}=M\overline{g}+\overline{R}$
• D. None of these

Answer 1

The correct option is A ${\overline{a}}_{CM}=\frac{M\overline{g}+\overline{R}}{M}$

before proceeding notice the following concept
1)Draw f.b.d of the rod.
2)acceleration of C.O.M in x-direction will be zero.
now we know that
$a_{cm}$=$\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}$
as we see that on the mass "m" two kind of forces will act which are "mg" and reaction "R"
so ${\overline{a}}_{cm}$=$\frac{m\overline{g}+\overline{R}}{m}$