Question

A uniform beam of length L and mass m is supported as shown. If the cable suddenly breaks, then

• A. the acceleration of end B is $\frac{9g}{7}\uparrow$
• B. the acceleration of end B is $\frac{9g}{7}\downarrow$
• C. the reaction at the pin support is $\frac{4mg}{7}$
• D. the reaction at the pin support is $\frac{2mg}{7}$

Answer 1

Answer: (B), (C)

the acceleration of end B is $\frac{9g}{7}\downarrow$
the reaction at the pin support is $\frac{4mg}{7}$
Let hinge point be P.
Torque about point ,

$\frac{3m}{4}\frac{3L}{8}-\frac{m}{4}\frac{L}{8}=I_p\alpha$ (clockwise)
Where $I_p$ is moment of inertia about point P.

$I_p=\frac{m{L}^2}{12}+m(\frac{L}{4}{)}^2=\frac{7m{L}^2}{48}$

Therefore,
$\frac{3m}{4}\frac{3L}{8}-\frac{m}{4}\frac{L}{8}=\frac{7m{L}^2}{48}\alpha$

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

$a_B=\alpha \frac{3L}{4}=\frac{9g}{7}$ in downward direction.

Now, from the frame of reference of the rod's center of mass.
$R\frac{L}{4}=I_c\alpha =\frac{m{L}^2}{12}\alpha$
$R=\frac{4mg}{7}$
Answer: B,C