JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 14)
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 :
_12th_January_Morning_Slot_en_14_1.png)
$${\overrightarrow r _{cm}}\, = {{11} \over 8}L\,\,\widehat x + {8 \over 8}L\widehat y$$
$${\overrightarrow r _{cm}}\, = {5 \over 8}L\,\,\widehat x + {{13} \over 8}L\widehat y$$
$${\overrightarrow r _{cm}}\, = {{13} \over 8}L\,\,\widehat x + {5 \over 8}L\widehat y$$
$${\overrightarrow r _{cm}}\, = {3 \over 8}L\,\,\widehat x + {{11} \over 8}L\widehat y$$
Explanation
_12th_January_Morning_Slot_en_14_2.png)
$${X_{cm}} = {{2mL + 2mL + {{5mL} \over 2}} \over {4m}} = {{13} \over 8}L$$
$${Y_{cm}} = {{2m \times L + m \times \left( {{L \over 2}} \right) + m \times 0} \over {4m}} = {{5L} \over 8}$$
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