question_answer

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) \[\frac{9}{20}mg\]

Correct Answer: A

Solution :

[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 , \[{{\alpha }_{CM}}=\alpha (AC)\] \[{{a}_{CM}}=\frac{3g}{2L}\left( \frac{9L}{20} \right)=\frac{27g}{40}\] Now , \[\frac{9}{10}mg-{{N}_{A}}=m{{a}_{CM}}=m.\frac{27g}{40}\] Or \[{{N}_{A}}=\frac{9}{40}mg\]