Question

A uniform rod of length $L$ (in between the supports) and mass m is placed on two hinges A and B. The rod breaks suddenly at length $\frac{L}{10}$ from the support B. Find the reaction at hinge A immediately after the rod breaks :

• A. $\frac{9}{40}\text{mg}$
• B. $\frac{19}{40}\text{mg}$
• C. $\frac{\text{mg}}{2}$
• D. $\frac{9}{20}\text{mg}$

Answer 1

The correct option is A $\frac{9}{40}\text{mg}$


Torque about A is $\tau =\frac{9}{10}mg\left(\frac{9}{20}L\right)=I\alpha =\frac{m}{3}{\left(\frac{9}{10}L\right)}^2\alpha$
$\alpha =\frac{3g}{2L}$
Acceleration,
$a_{CM}=\alpha \left(AC\right)=\frac{3g}{2L}\left(\frac{9L}{20}\right)=\frac{27g}{40}$
Now,
$\frac{9}{10}mg-N_A=m{a}_{CM}=m.\frac{27g}{40}$
$N_A=\frac{9}{40}\text{mg}$