Question 161: Centre of Mass and Rotational Motion

Question

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 :

Accepted Answer

$${x_{COM}} = {{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}} \over {{m_1} + {m_2} + {m_3}}}$$
= $${{0 + \left( {100} \right)10 + \left( {150} \right)\left( {0.5} \right)} \over {300}}$$
= $${{175} \over {300}}$$ = $${7 \over {12}}$$ 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

Answer

$$\left( {{{\sqrt 3 } \over 4}m,{5 \over {12}}m} \right)$$

Answer

$$\left( {{7 \over {12}}m,{{\sqrt 3 } \over 8}m} \right)$$

Answer

$$\left( {{7 \over {12}}m,{{\sqrt 3 } \over 4}m} \right)$$

Answer

$$\left( {{{\sqrt 3 } \over 8}m,{7 \over {12}}m} \right)$$

Centre of Mass and Rotational Motion

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 :

Choose the correct answer: