Physics, asked by muddy4852, 11 months ago

The moment of inertia of a hollow sphere of mass m and inner radius r and outer radius 2r having uniform mass distribution about diameter axis is

Answers

Answered by abhi178
1

Moment of inertia of hollow sphere is 62/35 Mr²

at first, let density of sphere is ρ.

volume of sphere of radius r, V = 4/3 πr³

mass of sphere of radius r, m = Vρ = 4/3πr³ρ

volume of sphere of radius 2r, V' = 4/3π(2r)³ = 32/3πr³

mass of sphere of radius 2r, m' = V'ρ

= 32/3πr³ρ

so, the mass of hollow sphere, M = 28/3 πr³ρ

⇒3M/28 = πr³ρ.......(1)

now, moment of inertia of hollow sphere = moment of inertia of bigger sphere (of radius 2r) - moment of inertia of smaller sphere ( of radius r)

= 2/5 m'(2r)² 2/5 mr²

[as we know, moment of inertia of sphere about an axis passing through its centre is 2/5 mr²]

= 2/5 r²[4m' - m]

= 2/5 r² [ 4 × 32/3 πr³ρ - 4/3 πr³ρ ]

= 8/15 r² × 31πr³ρ..........(2)

putting equation (1) in equation (2),

= 8/15 r² × 31(3M/28)

= 62/35 Mr²

hence, moment of inertia of hollow sphere is 62/35 Mr²

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