A hollow sphere has a mass of 15 kg, an inner radius of 12 cm and an outer radius of 18 cm. What is the rotational inertia (moment of inertia) of the sphere about an axis passing through its center

0.2396

Explanation:

Given that

Mass of the hollow sphere, m = 15 kg

Inner radius, r = 12 cm = 0.12 m

Outer radius, R = 18 cm = 0.18 m

Volume of a sphere is expressed as

V = 4/3.π.R³

Density ρ = mass / volume, therefore

Mass = ρ * v

Mass of the hollow sphere is given as

Mass of the outer sphere - mass of the inner sphere

M = ρV(o) - ρV(i)

V(o) = 4/3 * 3.142 * 0.18³

V(o) = 0.0244

V(i) = 4/3 * 3.142 * 0.12³

V(i) = 0.00724

15 = ρ (0.0244 - 0.00724)

15 = ρ (0.01716)

ρ = 15 / 0.01716

ρ = 874 kg/m³

moment of inertia about its centroidal axis is

I = 2/5 ρVR²

I(h) = I(o) - I(i)

I(h) = (2/5 * 874 * 0.0244 * 0.18²) - (2/5 * 874 * 0.00724 * 0.12²)

I(h) = 0.276 - 0.0364

I(h) = 0.2396

it is the second option.

answer: (431,2)

step-by-step explanation: