1.3 Solutions
37
Fig. 1.25
For continuous mass distribution
y CM =
1
M
y dm
Let σ be the surface density (mass per unit area), so that
M =
1
2
π R
2
σ
In polar coordinates dm = σ dA = σ r dθ dr
where dA is the element of area. Let the centre of mass be located at a distance
y CM from O along y-axis for reasons of symmetry:
y CM =
1
1
2 π R 2 σ
R
0
π
0
(r sin θ )(σ r dθ dr ) =
2
π R 2
R
0
r
2
π
0
sin θ dθ =
4R
3π
1.45 Let O be the origin, the centre of the base of the hemisphere, the z-axis being
perpendicular to the base. From symmetry the CM must lie on the z-axis,
Fig. 1.26. If ρ is the density, the mass element, dm = ρ dV , where dV is the
volume element:
Z CM =
1
M
Z dm =
1
M
Zρ dV
(1)
In polar coordinates, Z = r cos θ
(2)
dV = r
2 sin θ dθ dφdr
(3)
0 < r < R; 0 < θ <
π
2
;
0 < φ < 2π
The mass of the hemisphere
M = ρ
2
3
π R
3
(4)
Using (2), (3) and (4) in (1)