question_answer

A rod of length L is composed of a uniform length 1/2 L of wood whose mass is m^ and a uniform length 1/2 L of brass whose mass is \[{{m}_{b}}\]. The moment of inertia I of the rod about an axis perpendicular to the rod and through its centre is equal to

A)  \[({{m}_{w}}+{{m}_{b}}){{L}^{2}}/12\] 

B)  \[({{m}_{w}}+{{m}_{b}}){{L}^{2}}/6\]

C)  \[({{m}_{w}}+{{m}_{b}}){{L}^{2}}/3\]    

D)  \[({{m}_{w}}+{{m}_{b}}){{L}^{2}}/2\]

Correct Answer: A

Solution :

                Moment of inertia                 \[{{M}_{1}}={{m}_{w}}\times \frac{{{(L/2)}^{2}}}{3}={{m}_{w}}\frac{{{L}^{2}}}{12}\] Moment of inertia                 \[{{M}_{2}}={{m}_{b}}\times \frac{{{(L/2)}^{2}}}{3}={{m}_{b}}\times \frac{{{L}^{2}}}{12}\] Moment of inertia                 \[M={{M}_{1}}+{{M}_{2}}\]                 \[={{m}_{w}}\frac{{{L}^{2}}}{12}+{{m}_{b}}\times \frac{{{L}^{2}}}{12}\]                 \[=({{m}_{w}}={{m}_{b}})\frac{{{L}^{2}}}{12}\]