A composite rod made up of two rods AB and BC | vedclass

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(B) Let the length of each rod be $L$ and mass be $M$. The rods are joined at $B$. The center of mass of rod $AB$ is at a distance $L/2$ from $B$,and

Vedclass · Accepted Answer

It tilts down on the side of $AB$

Vedclass · Answer

(B) Let the length of each rod be $L$ and mass be $M$. The rods are joined at $B$. The center of mass of rod $AB$ is at a distance $L/2$ from $B$,and the center of mass of rod $BC$ is at a distance $L/2$ from $B$.When the rod is heated,the length of each rod increases according to the formula $\Delta L = L \alpha \Delta T$.Since $\alpha_{AB} > \alpha_{BC}$,the rod $AB$ expands more than the rod $BC$.Let the new lengths be $L_{AB}' = L(1 + \alpha_{AB} \Delta T)$ and $L_{BC}' = L(1 + \alpha_{BC} \Delta T)$.The new center of mass of rod $AB$ is at $L_{AB}'/2$ from $B$,and the new center of mass of rod $BC$ is at $L_{BC}'/2$ from $B$.Since $L_{AB}' > L_{BC}'$,the center of mass of rod $AB$ shifts further away from $B$ compared to the center of mass of rod $BC$.Because the rod is suspended at $B$,the torque due to the weight of rod $AB$ is $	au_{AB} = M g (L_{AB}'/2)$ and the torque due to the weight of rod $BC$ is $	au_{BC} = M g (L_{BC}'/2)$.Since $L_{AB}' > L_{BC}'$,the torque on the side of $AB$ is greater than the torque on the side of $BC$.Therefore,the rod tilts down on the side of $AB$.

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