Question

a uniform beam of length l and mass m is supported as shown if the cable at v suddenly breaks determime acceleration of end b and the reaction at the pin suport

Answer 1

Let hing point be P
now torque about point P,
$\frac{3mg}{4}\frac{3L}{8}-\frac{mg}{4}\frac{L}{8}=I_P\alpha$ (clockwise)
here $I_p$ is the moment of inertia about P.
$\frac{9mgL}{32}-\frac{mgL}{32}=I_P\alpha$

$\frac{mgL}{4}=I_P\alpha$.........(1)

so, $I_P=\frac{mL^2}{12}+m\left(\frac{L}{4}\right)^2$

$I_P=\frac{7mL^2}{48}$

so, $\frac{mgL}{4}=\frac{7mL^2}{48}$

$\alpha=\frac{12g}{7L}$

we know, $a_B=\alpha\left(\frac{3L}{4}\right)$

$a_B=\frac{9g}{7}$ in downward direction.

now normal reaction at the pin support is R
then, $R\left(\frac{3L}{4}\right)=I_c\alpha$

$R\left(\frac{L}{4}\right)=\frac{mL^2}{12}\frac{9g}{7}$

$R=\frac{4mg}{7}$