Question 187: Centre of Mass and Rotational Motion

Question

The position vector of the centre of mass $$\overrightarrow r {\,_{cm}}\,$$ of an asymmetric uniform bar of negligible area of crosssection as shown in figure is :

Accepted Answer

$${X_{cm}} = {{2mL + 2mL + {{5mL} \over 2}} \over {4m}} = {{13} \over 8}L$$
$${Y_{cm}} = {{2m 	imes L + m 	imes \left( {{L \over 2}} ight) + m 	imes 0} \over {4m}} = {{5L} \over 8}$$

Answer

$${\overrightarrow r _{cm}}\, = {{11} \over 8}L\,\,\widehat x + {8 \over 8}L\widehat y$$

Answer

$${\overrightarrow r _{cm}}\, = {5 \over 8}L\,\,\widehat x + {{13} \over 8}L\widehat y$$

Answer

$${\overrightarrow r _{cm}}\, = {{13} \over 8}L\,\,\widehat x + {5 \over 8}L\widehat y$$

Answer

$${\overrightarrow r _{cm}}\, = {3 \over 8}L\,\,\widehat x + {{11} \over 8}L\widehat y$$

Centre of Mass and Rotational Motion

The position vector of the centre of mass $\overrightarrow r {\,_{cm}}\,$ of an asymmetric uniform bar of negligible area of crosssection as shown in figure is :

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