Question

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

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

Answer 1

Answer: (A)

Torque $=\tau =\frac{9}{10}mg\left(\frac{9}{20}L\right)$
$=I\alpha =\frac{m}{3}{\left(\frac{9}{10}L\right)}^2\alpha$
$\alpha =\frac{3g}{2L}$
Acceleration, $a_{CM}=\alpha (AC)$

$a_{CM}=\frac{3g}{2L}\left(\frac{9L}{20}\right)=\frac{27g}{40}$
Now,$\frac{9}{10}mg-N_A=m{a}_{CM}=m\cdot \frac{27g}{40}$
or $N_A=\frac{9}{40}mg$