JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 23)
Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral triangle of side
1 m (as shown in the figure). The (x, y) coordinates of the centre of mass will be :
_12th_April_Evening_Slot_en_23_1.png)
$$\left( {{{\sqrt 3 } \over 4}m,{5 \over {12}}m} \right)$$
$$\left( {{7 \over {12}}m,{{\sqrt 3 } \over 8}m} \right)$$
$$\left( {{7 \over {12}}m,{{\sqrt 3 } \over 4}m} \right)$$
$$\left( {{{\sqrt 3 } \over 8}m,{7 \over {12}}m} \right)$$
Explanation
_12th_April_Evening_Slot_en_23_2.png)
$${x_{COM}} = {{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}} \over {{m_1} + {m_2} + {m_3}}}$$
= $${{0 + \left( {100} \right)10 + \left( {150} \right)\left( {0.5} \right)} \over {300}}$$
= $${{175} \over {300}}$$ = $${7 \over {12}}$$ m
$${y_{COM}} = {{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3}} \over {{m_1} + {m_2} + {m_3}}}$$
= $${{0 + \left( {100} \right)0 + \left( {150} \right){{\sqrt 3 } \over 2}} \over {300}}$$
= $${{75\sqrt 3 } \over {300}}$$ = $${{\sqrt 3 } \over 4}$$ m
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