Find the centre of mass of a hemispherical shell of radius R
Answers
Answer:
We are having a hollow hemisphere of mass M and radius R. The centre of mass of the hollow hemisphere will lie on the y-axis, which is the line passing through the centre of the base of the hollow hemisphere.
We are considering an elemental strip of width Rdθ and has a mass dM. The radius of the elemental ring is r = Rsinθ.
The elemental mass dM = (M/2πR2) (2πRsinθ.Rdθ)
dM = Msinθdθ
The y-coordinate of the centre of mass yc =(1/M)∫ydM
Putting the values of y and dM, we get
yc =(1/M)∫(Rcosθ)(Msinθdθ)
y_{c}=\frac{1}{M}\int_{0}^{\pi /2}(Rcos\theta )(Msin\theta d\theta )y
c
=
M
1
∫
0
π/2
(Rcosθ)(Msinθdθ)
y_{c}=\frac{1}{M}\int_{0}^{\pi /2}(Rcos\theta )(Msin\theta d\theta ) \times \frac{2}{2}y
c
=
M
1
∫
0
π/2
(Rcosθ)(Msinθdθ)×
2
2
y_{c}=\frac{R}{2}\int_{0}^{\pi /2}(sin2\theta d\theta )y
c
=
2
R
∫
0
π/2
(sin2θdθ)
= R/2(½ + ½) = R/2
Therefore, the centre of mass of a hollow hemisphere will be at R/2 along the y-axis. Here R is the radius of the hollow hemisphere.
Answer:
We are having a hollow hemisphere of mass M and radius R. The centre of mass of the hollow hemisphere will lie on the y-axis, which is the line passing through the centre of the base of the hollow hemisphere.
We are considering an elemental strip of width Rdθ and has a mass dM. The radius of the elemental ring is r = Rsinθ.
The elemental mass dM = (M/2πR2) (2πRsinθ.Rdθ)
dM = Msinθdθ
The y-coordinate of the centre of mass yc =(1/M)∫ydM
Putting the values of y and dM, we get
yc =(1/M)∫(Rcosθ)(Msinθdθ)
y_{c}=\frac{1}{M}\int_{0}^{\pi /2}(Rcos\theta )(Msin\theta d\theta )y
c
=
M
1
∫
0
π/2
(Rcosθ)(Msinθdθ)
y_{c}=\frac{1}{M}\int_{0}^{\pi /2}(Rcos\theta )(Msin\theta d\theta ) \times \frac{2}{2}y
c
=
M
1
∫
0
π/2
(Rcosθ)(Msinθdθ)×
2
2
y_{c}=\frac{R}{2}\int_{0}^{\pi /2}(sin2\theta d\theta )y
c
=
2
R
∫
0
π/2
(sin2θdθ)
= R/2(½ + ½) = R/2
Therefore, the centre of mass of a hollow hemisphere will be at R/2 along the y-axis. Here R is the radius of the hollow hemisphere.
Explanation:
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