Question

Find the center of mass of a uniform semicircular ring of radius r and mass m

Answer 1

Answer:

Answer

As the wire is uniform, the mass per unit length of the wire is$\dfrac{M}{\pi R}$

πR

M

. The mass of the element is, therefore,

d m =$\left( \dfrac { M } { \pi R } \right) ( R d \theta ) = \dfrac { M } { \pi } d \thetadm=( </p><p>πR$

M

)(Rdθ)=

π

M

dθ

The coordinates of the center of mass are

X =$\dfrac { 1 } { M } \int x d m = \dfrac { 1 } { M } \int _ { 0 } ^ { \pi } ( R \cos \theta ) \left( \dfrac { M } { \pi } \right) d \theta = 0X= </p><p>M</p><p>1</p><p> </p><p> ∫xdm= </p><p>M</p><p>1</p><p> </p><p> ∫ </p><p>0</p><p>π</p><p> </p><p> (Rcosθ)( </p><p>π</p><p>M</p><p> </p><p> )dθ=0</p><p>Y = \dfrac { 1 } { M } \int y d m = \dfrac { 1 } { M } \int _ { 0 } ^ { \pi } ( R \sin \theta ) \left( \dfrac { M } { \pi } \right) d \theta = \dfrac { 2 R } { \pi }Y= </p><p>M</p><p>1</p><p> </p><p> ∫ydm= </p><p>M</p><p>1</p><p> </p><p> ∫ </p><p>0</p><p>π</p><p> </p><p> (Rsinθ)( </p><p>π</p><p>M</p><p> </p><p> )dθ= </p><p>π</p><p>2R</p><p> </p><p> </p><p>Hence, position of center of mass, \left( 0,\,\dfrac{2R}{\pi } \right)(0, </p><p>π</p><p>2R</p><p> </p><p> )</p><p>solution$

Answer 2

Explanation:

As the wire is uniform, the mass per unit length of the wire is \dfrac{M}{\pi R}

πRM . The mass of the element is, therefore,

d m = \left( \dfrac { M } { \pi R } \right) ( R d \theta ) = \dfrac { M } { \pi } d \thetadm=(

πR

M

)(Rdθ)=

π

M

dθ

The coordinates of the center of mass are

X = \dfrac { 1 } { M } \int x d m = \dfrac { 1 } { M } \int _ { 0 } ^ { \pi } ( R \cos \theta ) \left( \dfrac { M } { \pi } \right) d \theta = 0X=

M

1

∫xdm=

M

1

∫

0

π

(Rcosθ)(

π

M

)dθ=0

Y = \dfrac { 1 } { M } \int y d m = \dfrac { 1 } { M } \int _ { 0 } ^ { \pi } ( R \sin \theta ) \left( \dfrac { M } { \pi } \right) d \theta = \dfrac { 2 R } { \pi }Y=

M

1

∫ydm=

M

1

∫

0

π

(Rsinθ)(

π

M

)dθ=

π

2R

Hence, position of center of mass, \left( 0,\,\dfrac{2R}{\pi } \right)(0,

π

2R

)