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 $
BETA
