Question

Mass per unit of a semicircular disc of total mass m radius R varies linearly with radial distance from the point $0$ (base centre). Find the height of centre of mass from $0$

• A. $\frac{2R}{\pi }$
• B. $\frac{3R}{2\pi }$
• C. $\frac{4R}{3\pi }$
• D. $\frac{3R}{8}$

Answer 1

The correct option is A $\frac{2R}{\pi }$

You have to need to me the fact that the centre of man of an uniform semi-circular circular ring of radius $r$ lies at a vertical distance of place on $\frac{2R}{\pi }$ axis on the centre. By symmetry the $x-$ coordinate of centre of man is $=0$

$x_{un}=0$

Now, we have to need to be certain things to find $Y_{un}$. At a vertical distance $Y$ from the origin imagine $u$ infinite i mal disk with a thickness $=dR$

The area of that infiniteterimal disk is $dA=\pi RdR$

let's find out it's mass,

$\Rightarrow dm=6dA$

$=\pi RdR\times {CR}^2$

$=\pi {CR}^{3}dR$.

$Y_r$ is the position of the centroid of that $dR$ ring and it equals $=\frac{2R}{\pi }$.