question_answer

The moment of inertia of a rod about an axis through its centre and perpendicular to it is \[\frac{1}{12}M{{L}^{2}}\](where M is the mass and L, the 12 length of the rod). The rod is bent in the middle so that the two halves make an angle of 60°. The moment of inertia of the bent rod about the same axis would be:

A) \[\frac{1}{48}M{{L}^{2}}\]

B)                       \[\frac{1}{12}M{{L}^{2}}\]           

C)       \[\frac{1}{24}M{{L}^{2}}\]           

D)       \[\frac{M{{L}^{2}}}{8\sqrt{3}}\]

Correct Answer: B

Solution :

Since, rod is bent at the middle, so each part of it will have same length \[\left( \frac{L}{2} \right)\] and mass \[\left( \frac{M}{2} \right)\] as shown.      Moment of inertia of each part through its one end \[=\frac{1}{3}\left( \frac{M}{2} \right){{\left( \frac{L}{2} \right)}^{2}}\] Hence, net moment of inertia through its middle point O is \[I=\frac{1}{3}\left( \frac{M}{2} \right){{\left( \frac{L}{2} \right)}^{2}}+\frac{1}{3}\left( \frac{M}{2} \right){{\left( \frac{L}{2} \right)}^{2}}\] \[=\frac{1}{3}\left[ \frac{M{{L}^{2}}}{8}+\frac{M{{L}^{2}}}{8} \right]=\frac{M{{L}^{2}}}{12}\]