Centre of mass of thin strip bent to form a semicircle of radius R is
Answers
Answer:
Centre of mass of a semi-circular ring of Radius R;
Calculation:
Linear mass density of ring
\lambda = \dfrac{m}{\pi R }λ=
πR
m
= > \dfrac{dm}{dy} = \dfrac{m}{\pi R }=>
dy
dm
=
πR
m
= > dm = \dfrac{m}{\pi R } \times (dy)=>dm=
πR
m
×(dy)
= > dm = \dfrac{m}{\pi R } \times (R \: d \theta)=>dm=
πR
m
×(Rdθ)
= > dm = \dfrac{m \: d \theta}{\pi }=>dm=
π
mdθ
So, centre of mass :
\displaystyle \: \bar{y} = \dfrac{1}{m} \int \: y \: dm
y
ˉ
=
m
1
∫ydm
= > \displaystyle \: \bar{y} = \dfrac{1}{m} \int \: R \cos( \theta)\: dm=>
y
ˉ
=
m
1
∫Rcos(θ)dm
= > \displaystyle \: \bar{y} = \dfrac{1}{m} \int \: R \cos( \theta)\: \frac{m \: d\theta}{\pi}=>
y
ˉ
=
m
1
∫Rcos(θ)
π
mdθ
= > \displaystyle \: \bar{y} = \dfrac{1}{\pi} \int \: R \cos( \theta) \: d \theta=>
y
ˉ
=
π
1
∫Rcos(θ)dθ
= > \displaystyle \: \bar{y} = \dfrac{R}{\pi} \int \: \cos( \theta) \: d \theta=>
y
ˉ
=
π
R
∫cos(θ)dθ
Putting limit ;
= > \displaystyle \: \bar{y} = \dfrac{R}{\pi} \int_{0}^{\pi} \: \cos( \theta) \: d \theta=>
y
ˉ
=
π
R
∫
0
π
cos(θ)dθ
= > \displaystyle \: \bar{y} = \dfrac{2R}{\pi}=>
y
ˉ
=
π
2R
So, final answer is:
\boxed{ \red{ \large{ \sf{ \: \bar{y} = \dfrac{2R}{\pi}}}}}
y
ˉ
=
π
2R