The coordinates of centre of mass of the following quarter circular arc is -

Question

The coordinates of centre of mass of the following quarter circular arc is - <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-m8c6qpkqwbek.png" /> (A) ((r/2) , (r/2)) (B) ((2 r/3) , (2 r/3)) (C) ((2 r/π ) , (2 r/π )) (D) ((4 r/π ) , (4 r/π ))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-m8c6qpkqt87a.png" /> dm=λ Rdθ =(2 m/π )dθ ∴ xc m =(1/m) displaystyle ∫ x d m =(1/m) displaystyle ∫ 0π / 2 R cos θ (2 m/π ) d θ Similarly yc m=(1/m) displaystyle ∫ 0(π /2) R sin . θ (2 m/π ) d θ

Q. The coordinates of centre of mass of the following quarter circular arc is -

$(\frac{r}{2}, \frac{r}{2})$

$(\frac{2 r}{3}, \frac{2 r}{3})$

$(\frac{2 r}{π}, \frac{2 r}{π})$

$(\frac{4 r}{π}, \frac{4 r}{π})$

$d m = λ R d θ = \frac{2 m}{π} d θ$
$∴ x_{c m} = \frac{1}{m} \int x d m$
$= \frac{1}{m} \int_{0}^{π /2} R cos θ \frac{2 m}{π} d θ$
Similarly $y_{c m} = \frac{1}{m} \int_{0}^{\frac{π}{2}} R s in . θ \frac{2 m}{π} d θ$

Q. The coordinates of centre of mass of the following quarter circular arc is -

(2r​,2r​)

(32r​,32r​)

(π2r​,π2r​)

(π4r​,π4r​)

dm=λRdθ=π2m​dθ ∴xcm​=m1​∫xdm =m1​∫0π/2​Rcosθπ2m​dθ Similarly ycm​=m1​∫02π​​Rsin.θπ2m​dθ

$(\frac{r}{2}, \frac{r}{2})$

$(\frac{2 r}{3}, \frac{2 r}{3})$

$(\frac{2 r}{π}, \frac{2 r}{π})$

$(\frac{4 r}{π}, \frac{4 r}{π})$

$d m = λ R d θ = \frac{2 m}{π} d θ$
$∴ x_{c m} = \frac{1}{m} \int x d m$
$= \frac{1}{m} \int_{0}^{π /2} R cos θ \frac{2 m}{π} d θ$
Similarly $y_{c m} = \frac{1}{m} \int_{0}^{\frac{π}{2}} R s in . θ \frac{2 m}{π} d θ$