Question

A rod of mass $M$ and length $L$ is placed in a horizontal plane with one end hinged about the vertical axis A horizontal force of $F=\frac{Mg}{2}$ is applied at a distance $\frac{5L}{6}$ from the hinged end . The angular acceleration of the rod will be :-

• A. $\frac{4g}{5L}$
• B. $\frac{5g}{4L}$
• C. $\frac{3g}{4L}$
• D. $\frac{4g}{3L}$

Answer 1

The correct option is B $\frac{5g}{4L}$

$T=I\alpha$

$T=\frac{M_g\times 5l}{2\times 6}=\frac{5Mgl}{12}$

$I=\frac{M{l}^2}{3}$

$\alpha =\frac{T}{I}=\frac{5mgl}{12}\times \frac{3}{{Ml}^2}=\frac{5g}{4l}$