Question

Find the position of the centre of mass of the T shaped plate from O in figure .

class 11

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Answer 1

Answer:

= 8multiply3multiply2 = 48

= 6 multiply 2=12

= 48 +12

Answer 2

★ Concept :-

Here the concept of centre of mass has been used. We see that we are given the dimensions of the figure. Firstly we will divide the figure into two parts and then find their area and then the centre of mass of both areas. The centre of mass of figure will lie between the centre of mass of both the parts. So by applying the value of centre of mass then, we can find the value of centre of mass of figure.

Let's do it !!

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★ Formula Used :-

$\;\boxed{\sf{\pink{CM\;=\;\bf{\dfrac{m_{1}x_{1}\:+\:m_{2}x_{2}\:+...\:+\:m_{n}x_{n}}{m_{1}\:+\:m_{2}\:+...\:+\:m_{n}}}}}}$

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★ Solution :-

Given,

» Length of Part I = 8 cm

» Breadth of Part I = 2 cm

» Length of Part II = 6 cm

» Breadth of Part II = 2 cm

Let,

• Centre of mass of figure = CM

• Centre of mass of Part I = O₁

• Centre of mass of Part II = O₂

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~ For the Centre of Mass of Part I and Part II ::

• For Part I :-

Here the distance of O₁ from O is 1 cm. Since both O₁ and O lie on same plane the x coordinate will be 0 . So,

»› Centre of mass of Part I = (0, 1)

• For Part II ::

Here the distance of O₂ from O is 5 cm (2 cm + 3 cm). Since both O₂ and O lie on same plane the x coordinate will be 0 . So,

»› Centre of mass of Part II = (0, 5)

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~ For the mass of Part I and Part II ::

We know that,

→ mass = Density × Volume

→ mass ∝ volume

Here since we are taking the dimensions in a plane. So,

→ mass ∝ area (m = gram since mass = g/cm³)

• For Part I :-

• Let the mass be m₁

→ m₁ ∝ length × breadth

(since the part is in rectangular shape)

→ m₁ ∝ 8 × 2

→ m₁ ∝ 16 g

• For Part II :-

• Let the mass be m₂

→ m₂ ∝ length × breadth

(since the part is in rectangular shape)

→ m₂ ∝ 16 × 2

→ m₂ ∝ 12 g

*Note : Here we will take mass as it is without taking constant to remove proportionality sign. This is because in the last part fraction that proportionality sign is going to be cancelled.

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~ For the Centre of Mass of Figure ::

We know that,

$\;\sf{\Longrightarrow\;\;CM\;=\;\bf{\dfrac{m_{1}x_{1}\:+\:m_{2}x_{2}\:+...\:+\:m_{n}x_{n}}{m_{1}\:+\:m_{2}\:+...\:+\:m_{n}}}}$

• Here x₁ = distance of O₁ from O = 1 cm

• m₁ = centre of mass of part I = 16 g

• Here x₂ = distance of O₂ from O = 5 cm

• m₂ = centre of mass of part II = 12 g

By applying values, we get

$\;\sf{\Longrightarrow\;\;CM\;=\;\bf{\dfrac{(16)(1)\:+\:(12)(5)}{16\:+\:12}}}$

$\;\sf{\Longrightarrow\;\;CM\;=\;\bf{\dfrac{16\:+\:60}{28}}}$

$\;\sf{\Longrightarrow\;\;CM\;=\;\bf{\dfrac{76}{28}}}$

$\;\sf{\Longrightarrow\;\;CM\;=\;\bf{2.7\;\;m}}$

(since 2.71 ≈ 2.7)

This is the required answer.

$\;\underline{\boxed{\tt{Required\;\:Centre\;\:of\;\:mass\;=\;\bf{\purple{2.7\;\:m}}}}}$

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★ More to know :-

$\;\tt{\leadsto\;\;K\;=\;\sqrt{\dfrac{I}{M}}}$

$\;\tt{\leadsto\;\;\tau\;=\;F\:\times\:d}$

$\;\tt{\leadsto\;\;L\;=\;I\:\times\:\omega}$

$\;\tt{\leadsto\;\;\tau\;=\;I\:\times\:\alpha}$

$\;\tt{\leadsto\;\;Rotational\;K.E.\;=\;\dfrac{1}{2}\:I\omega^{2}}$