Question

a solid sphere of radius r and mass m rotating about an axis passing through its center has a rotational inertia equal to 2/5MR². what is the rotational inertia of the sphere about an axis parallel to the original axis and passing through a point on its surface?

Answer 1

The rotational inertia of a solid sphere about an axis parallel to the one passing through its center and touching a point on its surface, using the parallel-axis theorem, is (7/5)mr².

Step-by-step explanation:

When determining the rotational inertia (or moment of inertia) of a solid sphere about an axis parallel to one passing through its center and touching a point on its surface, we must employ the parallel-axis theorem. The theorem states that the moment of inertia about the new axis (I_parallel) is equal to the moment of inertia about the central axis (I_center) plus the product of the mass of the sphere (m) and the square of the distance between the two axes (d^2). In this case, d equals the radius of the sphere (r).

Since the rotational inertia about the center is given by ¾ mr², using the parallel-axis theorem:

I_parallel = I_center + md²

I_parallel = (2/5)mr² + m(r²)

I_parallel = (2/5)mr² + (5/5)mr²

I_parallel = (7/5)mr²

Therefore, the rotational inertia of the sphere about an axis parallel to the original axis and passing through a point on its surface is (7/5)mr².