A) \[L/3\]
B) \[L/2\]
C) \[2L/3\]
D) \[3L/4\]
Correct Answer: B
Solution :
Let the mass of an element of length dx of the rod located at a distnace\[x\]away from left end is\[\frac{M}{L}dx\]. The\[x-\]coordinate of the centre of mass is given by \[{{X}_{cm}}=\frac{1}{M}\int{x\,dm}\] \[=\frac{1}{M}\int_{0}^{x}{x}\left( \frac{M}{L}dx \right)\] \[=\frac{1}{L}\left( \frac{{{x}^{2}}}{2} \right)_{0}^{L}=\frac{L}{2}\] The y-coordinate is \[{{Y}_{cm}}=\frac{1}{M}\int{ydm}=0\] and similarly, \[{{Z}_{cm}}=0\] Hence, the centre of mass is at\[\left( \frac{L}{2},0,0 \right)\]or at the middle point of the rod, i.e., at\[\frac{L}{2}\].You need to login to perform this action.
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