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## GR0877 Problem 84

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) The $x$-coordinate of the rod’s center of mass is $\displaystyle x_c = \frac{1}{M} \int_0^L x dm = \frac{1}{M} \int_0^L x \frac{2M}{L^2} xdx = \frac{2}{L^2} \int_0^L x^2 dx = \frac{2}{L^2} \cdot \frac{L^3}{3} = \frac{2L}{3}$.

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1. 11.11.2011 at 00:01

Also, you can eliminate A-D on the grounds that since the density increases with x, the CM will be displaced to the right of the CM were the rod density uniform.

• 25.10.2014 at 05:10

Also how I arrived at the correct answer. Cheers.