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?
You can find the appropriate formula here:
http://en.wikipedia.org/wiki/List_of_moments_of_inertia
To find the rotational inertia or moment of inertia of a hollow sphere, you can use the formula:
I = (2/3) * m * (r₁² + r₂²)
where
I = rotational inertia or moment of inertia
m = mass of the sphere
r₁ = inner radius of the sphere
r₂ = outer radius of the sphere
In this case, the mass of the sphere, m, is given as 15 kg, the inner radius, r₁, is 12 cm (or 0.12 m), and the outer radius, r₂, is 18 cm (or 0.18 m).
Substituting the given values into the formula, we have:
I = (2/3) * 15 kg * ((0.12 m)² + (0.18 m)²)
Simplifying the equation, we get:
I = (2/3) * 15 kg * (0.0144 m² + 0.0324 m²)
Next, we can perform the calculations:
I = (2/3) * 15 kg * (0.0468 m²)
Now, multiply the numbers together:
I = 9.36 kg * m²
Therefore, the rotational inertia or moment of inertia of the hollow sphere about an axis passing through its center is 9.36 kg * m².