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  4. 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.

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Q. 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.

System of Particles and Rotational Motion

Solution:

Torque $=\tau=\frac{9}{10} m g\left(\frac{9}{20} L\right)$
$=I \alpha=\frac{m}{3}\left(\frac{9}{10} L\right)^{2} \alpha$
$\alpha=\frac{3 g}{2 L}$
Acceleration, $a_{ CM }=\alpha(A C)$
image
$a_{ CM }=\frac{3 g}{2 L}\left(\frac{9 L}{20}\right)=\frac{27 g}{40}$
Now,$\frac{9}{10} m g-N_{A}=m a_{ CM }=m \cdot \frac{27 g}{40}$
or $N_{A}=\frac{9}{40} m g$