Question

(3,4) solve it... jaldi

yaad rakhna, spamm is not allowed
vo karega, uska sara answer, questions ka report kar dunga.

I need it. so please help me ​

Answer 1

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}$

Hi, friend

previously I had already solved this type of your question.

this question is also like them.

Do it yourself, please.

Again I write it.

$\Huge{\textbf{\textsf{{\purple{Ans}}{\pink{wer}}{\color{pink}{:}}}}}$

step 2:

Here, the centre of gravity G lies on the X- axis,

So,

$\bar y \: = 0$

Taking moments about the Y-axis,

$\bar x \sum \: w \: = \sum xw \:\:----(1)$

Conisder the lamina as made up of strips of area a parallel to the Y- axis;

Let the weight par unit area of an element be = p

so, w = p×a = pa

Now from (1) ⇨

$\displaystyle \bar x \sum \: pa\: = \sum xpa \\ \implies \: \bar xp \sum \: a = p \: \sum \: xa \\ \implies \: \bar x \sum \: a \: = \sum xa\: \\ \implies \: \bar x \sum \: 2y \delta x\: = \sum x(2y \delta x) \\ \\ now \: \sum \: = \: \int \: dx \\ \\ \displaystyle \bar x ×\int_0^4\:2y\:dx\:=\:\int_0^4\:2xy \: dx \: \: \: \: \: \{upper \: and \: lower \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: limit \: are \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: collected \: from \: the \: graph \} \\ \implies \: \displaystyle \bar x ×\int_0^4\:2 \times 3 {x}^{ \frac{1}{2} } \:dx\:=\:\int_0^4\:2x \times {x}^{ \frac{1}{2} } \: dx \\ \implies \: \displaystyle \bar x ×\int_0^4\:6 {x}^{ \frac{1}{2} } \:dx\:=\:\int_0^4\:6 {x}^{ \frac{3}{2} } \: dx \\ \implies \: \bar x \times 32 = \frac{384}{5} \\ \implies \: \bar x = \frac{12}{5}$

$So,\: G\:=\:(\:\frac{12}{5}\:,\:0\:)$

Answer 2

Answer:

A

1

= area of complete circle = πa

2

A

2

= area of small circle = π(

2

a

)

2

=

4

πa

2

(x

1

,y

1

) = coordinates of center of mass of large circle = (0, 0)

(x

2

,y

2

) = coordinates of center of mass of small circle = (

2

a

,0)

Using x

COM

=

A

1

−A

2

A

1

x

1

−A

2

x

2

we get x

COM

=

πa

2

−

4

π

2

−

4

πa

2

(

2

a

)

=

(

4

3

)

−(

8

1

)

a=−

6

a

and get x

COM

=0 as y

1

and y

1

both are zero.

Therefore, coordinates of COM of the lamina shown in figure are (−

6

a

,0).

Step-by-step explanation:

please give many thanks