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?
To find the rotational inertia of a hollow sphere about an axis passing through its center, we 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
Given:
m = 15 kg
r₁ = 12 cm
r₂ = 18 cm
First, we need to convert the radius from centimeters to meters since the mass is given in kilograms.
1 cm = 0.01 m
So, the inner radius(r₁) in meters is:
r₁ = 12 cm * 0.01 m/cm = 0.12 m
And the outer radius(r₂) in meters is:
r₂ = 18 cm * 0.01 m/cm = 0.18 m
Now we can substitute the values into the formula:
I = 2/3 * 15 kg * (0.12 m)² + (0.18 m)²
Calculating this will give us the value of the rotational inertia of the hollow sphere about the given axis.