Rotational Motion Question 139

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.

Options:

A) $\frac{9}{40}mg$

B) $\frac{19}{40}mg$

C) $\frac{mg}{2}$

D) $\frac{9}{20}mg$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] 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} $

$\text { Acceleration, } \alpha _{C M}=\alpha(A C) $

$ a _{C M}=\frac{3 g}{2 L}\left(\frac{9 L}{20}\right)=\frac{27 g}{40} $

$ \text { Now, } \frac{9}{10} m g-N_A=m a _{C M}=m \cdot \frac{27 g}{40} \text { Or } $

$ N_A=\frac{9}{40} m g $



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