Four cubes of side “a” each of mass 40gm, 20gm, 10gm and 20gm are arranged in XY plane as shown in f

Question

Four cubes of side “a” each of mass 40gm, 20gm, 10gm and 20gm are arranged in XY plane as shown in figure. The coordinates of centre of mass of the combination with respect to O, are :

Accepted Answer

Sol. Block Its Com 40 gm $$\left( \frac{a}{2},\mspace{6mu}\frac{3a}{2} \right)$$ 20 gm $$\left( \frac{a}{2},\mspace{6mu}\frac{a}{2} \right)$$ 10 gm $$\left( \frac{3a}{2},\mspace{6mu}\frac{a}{2} \right)$$ 20 gm $$\left( \frac{5a}{2},\mspace{6mu}\frac{a}{2} \right)$$ $\overrightarrow{r_{cm}} = \frac{\sum_{}^{}{m_{i}\overrightarrow{r_{i}}}}{\sum_{}^{}m_{i}}$ = $\frac{40\left( \frac{a}{2},\frac{3a}{2} \right) + 20\left( \frac{a}{2},\frac{a}{2} \right) + 10\left( \frac{3a}{2},\frac{a}{2} \right) + 20\left( \frac{5a}{2},\frac{a}{2} \right)}{40 + 20 + 10 + 20}$ = $\left( \frac{19}{18}a,\mspace{6mu}\frac{17}{18}a \right)$

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More Center of Mass Questions

Block	Its Com
40 gm	$$\left( \frac{a}{2},\mspace{6mu}\frac{3a}{2} \right)$$
20 gm	$$\left( \frac{a}{2},\mspace{6mu}\frac{a}{2} \right)$$
10 gm	$$\left( \frac{3a}{2},\mspace{6mu}\frac{a}{2} \right)$$
20 gm	$$\left( \frac{5a}{2},\mspace{6mu}\frac{a}{2} \right)$$