Question

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class 11.. physics

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Answer 1

Answer:

Center of mass is given by=

(m1x1 + m2x2 + m3x3 +m4x4)

—————————————————

(m1 +m2+m3+m4)

The diagonal of a square is 0.8 meters

therefore using pythagoras law each side is of a distance 0.56 meters

So centre of mass

8×0×2×0.56+9×0.8+2×0.56

————--------------------------------

8+2+4

=0.32meter

=30cm

So (B) is correct answer

♥jao Di soo jao abb Good night sweet dreams take care bubye♥

Answer 2

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

Here the Concept of Centre of Mass has been used. We see that we are given the Diagonal of the Square. So first we need to find the side of the square. This can be done by Pythagoras Theorem . Also for calculating the distance between the mass, we need to find the closest distance. The reason because we need to calculate distance of Centre of Mass from A. So we need to find closest distance of points from A.

After finding all these values, we can apply in formula and find our answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}}$

$\\\;\boxed{\sf{\vec{R}_{(CM)}\;=\;\bf{\dfrac{m_{1}x_{1}\;+\;m_{2}x_{2}\;+\;m_{3}x_{3}\;+\;m_{4}x_{4}}{m_{1}\;+\;m_{2}\;+\;m_{3}\;+\;m_{4}}}}}$

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★ Solution :-

Given,

» Diagonal of the Square = 80 cm = 0.8 m

» Mass at the point A = 8 Kg

» Mass at the point B = 2 Kg

» Mass at the point C = 4 Kg

» Mass at the point D = 2 Kg

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~ For the Side of the Square :-

We know that every angle of Square = 90°

Then from figure, ∆ABC is a Right - Angled Triangle. Also all sides of Square are equal.

Let the each side of square be 'x'. Then,

✒ Hypotenuse = AC = 0.8 m

✒ Base = AB = x

✒ Height = BC = AB = x

Then, by Pythagoras Theorem, we get,

$\\\;\;\sf{:\rightarrow\;\;(Hypotenuse)^{2}\;=\;\bf{(Base)^{2}\;+\;(Height)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(0.8)^{2}\;=\;\bf{(x)^{2}\;+\;(x)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;(0.64)^{2}\;=\;\bf{2\:x^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;x^{2}\;=\;\bf{\dfrac{0.64}{2}}}$

$\\\;\;\sf{:\rightarrow\;\;x^{2}\;=\;\bf{0.32}}$

$\\\;\;\sf{:\rightarrow\;\;x\;=\;\bf{\sqrt{0.32}}}$

$\\\;\;\underline{\underline{\bf{:\rightarrow\;\;x\;=\;\bf{0.56\;\;m}}}}$

Hence, AB = BC = CD = DA = 0.56 m

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~ For the Centre of Mass from A ::

For this, we need to get closest distance of different points from A.

Here, R denotes the distance of Centre of Mass from point A.

$\\\;\;\sf{:\Rightarrow\;\;For\;point\;A\;:\;m_{1}\;=\;8\;Kg\;\;,\;\;x_{1}\;=\;0\;m}$

• This is because at point A, distance of point A will be zero only.

$\\\;\;\sf{:\Rightarrow\;\;For\;point\;B\;:\;m_{2}\;=\;2\;Kg\;\;,\;\;x_{2}\;=\;0.56\;m}$

• The nearest distance of point B from A is the side of square.

$\\\;\;\sf{:\Rightarrow\;\;For\;point\;C\;:\;m_{3}\;=\;4\;Kg\;\;,\;\;x_{3}\;=\;0.8\;m}$

• The nearest distance of point C from A is the diagonal of the square.

$\\\;\;\sf{:\Rightarrow\;\;For\;point\;D\;:\;m_{4}\;=\;2\;Kg\;\;,\;\;x_{4}\;=\;0.56\;m}$

• The nearest distance of point D from A is the side of the square.

Then applying these values in the formula, we get,

$\\\;\;\sf{:\Longrightarrow\;\;\vec{R}_{(CM)}\;=\;\bf{\dfrac{(m_{1}x_{1})\;+\;(m_{2}x_{2})\;+\;(m_{3}x_{3})\;+\;(m_{4}x_{4})}{m_{1}\;+\;m_{2}\;+\;m_{3}\;+\;m_{4}}}}$

$\\\;\;\sf{:\Longrightarrow\;\;\vec{R}_{(CM)}\;=\;\bf{\dfrac{(8\;\times\;0)\;+\;(2\;\times\;0.56)\;+\;(4\;\times\;0.8)\;+\;(2\;\times\;0.56)}{8\;+\;2\;+\;4\;+\;2}}}$

$\\\;\;\sf{:\Longrightarrow\;\;\vec{R}_{(CM)}\;=\;\bf{\dfrac{0\;+\;1.12\;+\;3.2\;+\;1.12}{16}}}$

$\\\;\;\sf{:\Longrightarrow\;\;\vec{R}_{(CM)}\;=\;\bf{\dfrac{5.44}{16}}}$

$\\\;\;\sf{:\Longrightarrow\;\;\vec{R}_{(CM)}\;=\;\bf{0.34\;\;m}}$

$\\\;\;\sf{:\mapsto\;\;\vec{R}_{(CM)}\;=\;\bf{0.30\;\;m}}$

This value is after rounding off the decimal digit.

$\\\;\;\bf{:\mapsto\;\;\vec{R}_{(CM)}\;=\;\bf{30\;\;cm}}$

So, option b.) 30 cm is correct option.

$\\\;\underline{\boxed{\tt{Hence,\;\;the\;\;required\;\;answer\;=\;\bf{30\;\;cm}}}}$

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★ More to know :-

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

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

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

$\\\;\sf{\leadsto\;\;Radius\;of\;Gyration,\;K\;=\;\sqrt{\dfrac{Moment\;of\;Inertia}{Mass}}}$

$\\\;\sf{\leadsto\;\;Angular\;Momentum\;=\;Linear\;Momentum\;\times\;Perpendicular\;Distance}$