Question

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

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Answer 1

Answer:

Given:

A quarter ring has been provided with centre at origin.

To find:

Coordinates of centre of mass at C

Concept:

For any sector subtending an angle of 2α at the centre, the distance of COM from the centre (0,0) is given by

$d = \frac{r \: sin( \alpha )}{ \alpha }$

" α" has to be taken in radians.

Calculation:

In this case;

∴ 2α = π/2

=> α = π/4

Putting the value of α in the Equation :

$= > d = \frac{r \: sin( \frac{\pi}{4} )}{ (\frac{\pi}{4} )}$

$= > d = \frac{4r}{ \sqrt{2} \pi}$

$= > d = \frac{2r \sqrt{2} }{\pi}$

Now, in this case, COM will be present along the Angle bisector (π/4) at a distance of d = (2r√2)/π from (0,0).

So, the coordinates can now be found with trigonometry.

$(x,y) \: = ( \frac{2r}{\pi} , \frac{2r}{\pi} )$

The coordinates come as :

Answer 2

Answer:

Given:

A quarter ring has been provided with centre at origin.

To find:

Coordinates of centre of mass at C

Concept:

For any sector subtending an angle of 2α at the centre, the distance of COM from the centre (0,0) is given by

\begin{lgathered}d = \frac{r \: sin( \alpha )}{ \alpha } \\\end{lgathered}d=αrsin(α)

" α" has to be taken in radians.

Calculation:

In this case;

∴ 2α = π/2

=> α = π/4

Putting the value of α in the Equation :

\begin{lgathered}= > d = \frac{r \: sin( \frac{\pi}{4} )}{ (\frac{\pi}{4} )} \\\end{lgathered}=>d=(4π)rsin(4π)

\begin{lgathered}= > d = \frac{4r}{ \sqrt{2} \pi} \\\end{lgathered}=>d=2π4r

\begin{lgathered}= > d = \frac{2r \sqrt{2} }{\pi} \\\end{lgathered}=>d=π2r2

Now, in this case, COM will be present along the Angle bisector (π/4) at a distance of d = (2r√2)/π from (0,0).

So, the coordinates can now be found with trigonometry.

\begin{lgathered}(x,y) \: = ( \frac{2r}{\pi} , \frac{2r}{\pi} ) \\\end{lgathered}(x,y)=(π2r,π2r)

The coordinates come as :