Question

The thin rod shown below has mass $M$ and length $L$. A force $F$ acts at one end as shown and the rod is free to rotate about the other end in horizontal plane. Initial angular acceleration of the rod is:

• A. $\frac{3F}{2ML}$
• B. $\frac{2F}{3ML}$
• C. $\frac{F}{ML}$
• D. $\frac{F}{2ML}$

Answer 1

The correct option is A $\frac{3F}{2ML}$

$r$ is perpendicular from the hinge point (axis)

On the line of action of the force

$\sin {30}^o-\frac{r}{hypotenuse}=\frac{r}{L}$

$r=L\sin {30}^o=\frac{L}{2}$

initial torque $\tau =Fr=\frac{FL}{2}$

moment of interita of rod w.r.t end axis will be $I=\frac{m{L}^2}{3}$

so initial angular acceleration $\alpha =\frac{\tau }{I}$

so $\alpha =\frac{FL/2}{m{L}^2/3}=\frac{3F}{2mL}$