center of mass of a solid hemisphere of radius R is 3R / 8 and the Mass of the solid is M.
An elemental disc at a height of h from the base of the hemisphere is taken.
The mass of the elemental disc is taken dM
and the width is taken dy.
et the solid hemisphere be of mass M and has the radius R.
The center of mass will lie vertically which passes through the center of the hemisphere.
Taking the elemental disc at a height h from the base of the hemisphere’s base. The DM is the mass of the elemental disc and the width is dy.
the radius of the disc is:
R = R²- y²
Mass of the disc dM
= (3 M/ 2π R³) × ( πr² dy ) …..(2)
eq.(1) in eq.(2)..................
dM = ( 3M / 2 π R³ ) × π ( R² – y² ) dy)
Y - coordinate of Centre of mass,
Yc = (1 / M) ഽ y dM
Here y is the y - coordinate representing the height of the elemental disc from the base.
Putting the dM and calculating the center of mass, we get
yc= (1/M) ഽ y(3M /2 R^3) × ( R^2 - y^2) dy)
Integrating between 0⟶R
yc = (1/M) ഽ y (3M / 2 R^3) × (R^2 - y^2) dy)
yc = ( 3/2r^3 ) ഽ × (R^3 y- y^3 ) dy
= ( 3/ 2R2)[ (R4 / 2 ) – ( R4 / 4)] = 3R / 8
yc = 3 R /8
Centre of Mass of solid hemisphere (yc) = 3 R / 8.
Mass of solid = M
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