Question

Find coordinates of the center of mass of a quarter ring of radius placed in the first quadrant of a Cartesian coordinate system, with center at the origin.

Answer 1

Hint: Quarter ring indicates that we need to find the center of mass for one-fourth of the ring placed in the first quadrant. We will find the general expression of coordinates of a small part of the ring. Finally, we will integrate it to find the required coordinates using the basic formula.

Formula Used:
$X_{CM}={\displaystyle \frac{1}{M}}\int xdm$
$Y_{CM}={\displaystyle \frac{1}{M}}\int ydm$

Complete step-by-step solution:
Here, we need to find the coordinates of the centre of mass of one fourth of a ring placed in the first quadrant. We will assume the origin as the centre of the circle, of which the ring is a part of, and then we can draw the diagram as follows:

Here, consider a one fourth ring present in the first quadrant having a total mass $M$ and radius $R$ and subtending a total angle of ${\displaystyle \frac{\pi }{2}}$ at the origin.
To find the centre of mass, consider a small portion of the ring at an angle $\theta$ from the x-axis. Let the mass of this portion be $dm$ of length $l$ and the angle subtended by this portion at the origin be $d\theta$.
Here, we will find the $x$ as well as $y$ component of the centre of mass and hence the coordinates of centre of mass.
Here, the total mass $M$ subtends an angle of ${\displaystyle \frac{\pi }{2}}$. Also, the mass $dm$ subtends an angle of $d\theta$
Hence, the mass $dm$can be written as:
$dm={\displaystyle \frac{M}{\left(\pi /2\right)}}\times d\theta$ ----(i)
Also, the length of the mass $dm$ can be written as:
$d\theta ={\displaystyle \frac{l}{R}}$
$\Rightarrow l=Rd\theta$ ----(ii)
Also, at any angle $\theta$ the general x coordinate and y coordinate can be written as:
$x=R\cos \theta$
$y=R\sin \theta$
For x coordinate:
$X_{CM}={\displaystyle \frac{1}{M}}\int xdm$
$\Rightarrow X_{CM}={\displaystyle \frac{1}{M}}\int R\cos \theta dm$
From equation (i)
$\Rightarrow X_{CM}={\displaystyle \frac{1}{M}}\int R\cos \theta {\displaystyle \frac{M}{(\pi /2)}}\times d\theta$
$\Rightarrow X_{CM}={\displaystyle \frac{2R}{\pi }}\underset{0}{\overset{\pi /2}{\int }}\cos \theta d\theta$
$\Rightarrow X_{CM}={\displaystyle \frac{2R}{\pi }}$
For y coordinate:
$Y_{CM}={\displaystyle \frac{1}{M}}\int ydm$
$\Rightarrow Y_{CM}={\displaystyle \frac{1}{M}}\int R\sin \theta dm$
From equation (i)
$Y_{CM}={\displaystyle \frac{1}{M}}\int R\sin \theta {\displaystyle \frac{M}{(\pi /2)}}d\theta$
$\Rightarrow Y_{CM}={\displaystyle \frac{2R}{\pi }}\underset{0}{\overset{\pi /2}{\int }}\sin \theta d\theta$
$\Rightarrow Y_{CM}={\displaystyle \frac{2R}{\pi }}$
Hence, the coordinates of the centre of mass of a quarter ring placed in the first quadrant is $(X_{CM},Y_{CM})=({\displaystyle \frac{2R}{\pi }},{\displaystyle \frac{2R}{\pi }})$.

Note: To find centre of mass, always take a small mass into consideration. Find out the general equation for the x and y coordinate for that small part. Then use the basic formula to find the coordinates of the centre of mass.