Question

Three points with masses $\sf m_1$ , $\sf m_2$ and $\sf m_3$ will be placed at the corners of a thin massless rectangular sheet (1.2 m × 1.0 m) as shown. Centre of mass will be located at the point ?​

Answer 1

• Refer to the attachment for sólution ✔️

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept Centre of Mass has been used. We need to find the point at which the Centre of Mass will be located. This means we need to find the coordinates of this distance. This distance will be given by x and y coordinates representing the centre of mass. We already know the formula. We can apply the values and find the coordinates.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{s_{(CM)}\;=\;\bf{\dfrac{m_{1}\:s_{1}\;+\;m_{2}\:s_{2}\;+\;m_{3}\:s_{3}}{m_{1}\;+\;m_{2}\;+\;m_{3}}}}}$

Here s shows the position of centre of mass, and s with subscripts shows position of individual mass at respective axis.

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★ Solution :-

Given,

» Dimensions of Rectangle = 1.2 m × 1.0 m

» Mass at point C = m₃ = 2.4 Kg

» Mass at point A = m₁ = 1.6 Kg

» Mass at point B = m₂ = 2.0 Kg

From Figure,

• Coordinates of point A = (0, 0)

(since, A lies at origin)

• Coordinates of point B = (1.2, 0)

(since, B lies on x - axis)

• Coordinates of point C = (0, 1)

(since, C lies on y - axis)

* CM denotes the Centre of Mass.

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~ For the Coordinate of Centre of Mass at X - Axis ::

By using the formula of Centre of Mass,

$\\\;\sf{:\rightarrow\;\;s_{(CM)}\;=\;\bf{\dfrac{m_{1}\:s_{1}\;+\;m_{2}\:s_{2}\;+\;m_{3}\:s_{3}}{m_{1}\;+\;m_{2}\;+\;m_{3}}}}$

when the values of x axis are applied to it, we get,

$\\\;\sf{:\Longrightarrow\;\;x_{(CM)}\;=\;\bf{\dfrac{m_{1}\:x_{1}\;+\;m_{2}\:x_{2}\;+\;m_{3}\:x_{3}}{m_{1}\;+\;m_{2}\;+\;m_{3}}}}$

• Point A :: m₁ = 1.6 Kg and x₁ = Coordinate of A on x - axis = 0

• Point B :: m₂ = 2 Kg and x₂ = Coordinate of B on x - axis = 1.2

• Point C :: m₃ = 2.4 Kg and x₃ = Coordinate of C on x - axis = 0

By applying these values in the formula, we get,

$\\\;\sf{:\Longrightarrow\;\;x_{(CM)}\;=\;\bf{\dfrac{(1.6)\:(0)\;+\;(2)\:(1.2)\;+\;(2.4)\:(0)}{1.6\;+\;2\;+\;2.4}}}$

$\\\;\sf{:\Longrightarrow\;\;x_{(CM)}\;=\;\bf{\dfrac{0\;+\;2.4\;+\;0}{6}}}$

$\\\;\sf{:\Longrightarrow\;\;x_{(CM)}\;=\;\bf{\dfrac{2.4}{6}}}$

$\\\;\bf{:\Longrightarrow\;\;x_{(CM)}\;=\;\bf{\green{0.4}}}$

$\\\;\underline{\boxed{\tt{Coordinate\;\;of\;\;CM\;\;on\;\;x\;-\;axis\;=\;\bf{\red{0.4}}}}}$

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~ For the Coordinate of Centre of Mass on Y - Axis ::

By using the formula of Centre of Mass, we get,

$\\\;\sf{:\rightarrow\;\;s_{(CM)}\;=\;\bf{\dfrac{m_{1}\:s_{1}\;+\;m_{2}\:s_{2}\;+\;m_{3}\:s_{3}}{m_{1}\;+\;m_{2}\;+\;m_{3}}}}$

when the values of y are applied to it, we get,

$\\\;\sf{:\Longrightarrow\;\;y_{(CM)}\;=\;\bf{\dfrac{m_{1}\:y_{1}\;+\;m_{2}\:y_{2}\;+\;m_{3}\:y_{3}}{m_{1}\;+\;m_{2}\;+\;m_{3}}}}$

• Point A :: m₁ = 1.6 Kg and y₁ = Coordinate of A on y - axis = 0

• Point B :: m₂ = 2 Kg and y₂ = Coordinate of B on y - axis = 0

• Point C :: m₃ = 2.4 Kg and y₃ = Coordinate of C on y - axis = 1

By applying these values in the formula, we get,

$\\\;\sf{:\Longrightarrow\;\;y_{(CM)}\;=\;\bf{\dfrac{(1.6)\:(0)\;+\;(2)\:(0)\;+\;(2.4)\:(1)}{1.6\;+\;2\;+\;2.4}}}$

$\\\;\sf{:\Longrightarrow\;\;y_{(CM)}\;=\;\bf{\dfrac{0\;+\;0\;+\;2.4}{6}}}$

$\\\;\sf{:\Longrightarrow\;\;y_{(CM)}\;=\;\bf{\dfrac{2.4}{6}}}$

$\\\;\bf{:\Longrightarrow\;\;y_{(CM)}\;=\;\bf{\green{0.4}}}$

$\\\;\underline{\boxed{\tt{Coordinate\;\;of\;\;CM\;\;on\;\;y\;-\;axis\;=\;\bf{\red{0.4}}}}}$

Now combining the x and y coordinates, we get,

✒ Position of Centre of Mass = (0.4, 0.4)

$\\\;\displaystyle{\underline{\underline{\rm{Thus,\;Position\;of\;CM\;is\;\;\boxed{\bf{\blue{(0.4,\:0.4)}}}}}}}$

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★ More Formulas to know :-

$\\\;\sf{\leadsto\;\;\omega\;=\;\omega_{0}\;+\;\alpha t}$

$\\\;\sf{\leadsto\;\;\theta\;=\;\omega_{0}t\;+\;\dfrac{1}{2}\alpha t^{2}}$

$\\\;\sf{\leadsto\;\;\omega^{2}\;=\;\omega_{0} ^{2}\;+\;2\alpha\theta}$

$\\\;\sf{\leadsto\;\;L\;=\;I\omega}$

$\\\;\sf{\leadsto\;\;\tau\;=\;I\alpha}$