The radius of gyration, k, of a body is defined as the square root of the ratio of the body's moment of inertia to its mass. For a solid sphere rotating about an axis passing through its center, the moment of inertia, I, is given by (2/5)MR^2, where M is the mass of the sphere and R is its radius.

However, in this case, the sphere is rotating about a tangent, which means the axis of rotation is at the edge of the sphere, not its center. This situation is equivalent to the parallel axis theorem, which states that the moment of inertia about an axis parallel to and a distance d away from an axis through the center of mass is given by I = I_cm + Md^2, where I_cm is the moment of inertia about the center of mass axis.

Here, d = R (since the axis is a distance R away from the center), and I_cm = (2/5)MR^2. Substituting these values in, we get I = (2/5)MR^2 + MR^2 = (7/5)MR^2.

The radius of gyration, k, is then given by sqrt(I/M) = sqrt((7/5)R^2) = sqrt(7/5) * R.

Question

The radius of gyration, k, of a body is defined as the square root of the ratio of the body's moment of inertia to its mass. For a solid sphere rotating about an axis passing through its center, the moment of inertia, I, is given by (2/5)MR^2, where M is the mass of the sphere and R is its radius.

However, in this case, the sphere is rotating about a tangent, which means the axis of rotation is at the edge of the sphere, not its center. This situation is equivalent to the parallel axis theorem, which states that the moment of inertia about an axis parallel to and a distance d away from an axis through the center of mass is given by I = I_cm + Md^2, where I_cm is the moment of inertia about the center of mass axis.

Here, d = R (since the axis is a distance R away from the center), and I_cm = (2/5)MR^2. Substituting these values in, we get I = (2/5)MR^2 + MR^2 = (7/5)MR^2.

The radius of gyration, k, is then given by sqrt(I/M) = sqrt((7/5)R^2) = sqrt(7/5) * R.

Knowee AI · Accepted Answer

The radius of gyration, k, of a body is defined as the square root of the ratio of the body's moment of inertia to its mass. For a solid sphere rotating about an axis passing through its center, the moment of inertia, I, is given by (2/5)MR^2, where M is the mass of the sphere and R is its radius.

However, in this case, the sphere is rotating about a tangent, which means the axis of rotation is at the edge of the sphere, not its center. This situation is equivalent to the parallel axis theorem, which states that the moment of inertia about an axis parallel to and a distance d away from an axis through the center of mass is given by I = I_cm + Md^2, where I_cm is the moment of inertia about the center of mass axis.

Here, d = R (since the axis is a distance R away from the center), and I_cm = (2/5)MR^2. Substituting these values in, we get I = (2/5)MR^2 + MR^2 = (7/5)MR^2.

The radius of gyration, k, is then given by sqrt(I/M) = sqrt((7/5)R^2) = sqrt(7/5) * R.

The radius of gyration of a sphere of radius R about a tangent is :-

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