Question

a solid circular cone of radius R is joined to a uniform solid hemisphere of radius R both are made of same material the centre of mass of the ball composite solid lies at the common based find the height of cone ​

Answer 1

h = √3r

Explanation:

The volume of the cone  = 1/3 π r (3) h

M1 is the Mass of the cone  = p x 1/3 π r (2) h

M2 is the mass of the hemisphere = p x 1/2 x 4/3 π r (3) = 2/3 π r (3)

Now as the Y =( m1 y1 + m2 y2) / m1 + m2

O =  [p x 1/3 π r (2) h x h/4 + p x 2/3 π r (3) (-3r/8) ] / p x 1/3 π r (2) h x h/4 + p x 2/3 π r (3)

Or it can be

p x 1/3 π r (2) [ h (2) /4 -2r x 3r/8 ]

0r

h = √3r

Answer 2

h = √3.r

Explanation:

Volume of cone  = 1/3πr²h

Mass of cone, m1 = ρ *  1/3 πr²h

Mass of hemisphere = ρ *  1/2 * 4/3 πr³ = ρ * 2/3 πr³

We know that Y = (m1 y1 + m2 y2) / (m1 + m2)

Substituting the values, we get:

0  = [ ρ *  1/3 πr²h * h/4] + [ ρ * 2/3 πr³ - 3r/8] / (ρ *  1/3 πr²h) + (ρ * 2/3 πr³)

ρ *  1/3 πr² [ h²/4 - 2r *3r/8] = 0

This implies that [ h²/4 - 2r *3r/8] = 0

h²/4 - 3r²/4 = 0

h²/4 = 3r²/4

h² = 3r²

Therefore h = √(3r²)

h = √3.r