Question

A solid sphere and a ring have equal masses and equal radius of gyration. If the sphere is rotating about its diameter and ring about an axis passing through and perpendicular to its plane, then the ratio of radius is \(\sqrt{\frac{x}{2} }\) then find the value of x.

Answer 1

Correct Answer: 5

When comparing a solid sphere rotating about its diameter to a ring rotating about an axis perpendicular to its plane, both have equal masses (M) and equal radii of gyration (K). The formula for the radius of gyration is K = √(I/M), where I is the moment of inertia.

For a solid sphere rotating about its diameter, the moment of inertia is Isphere = (2/5)MR2. Therefore, the radius of gyration is:

K_sphere = √((2/5)MR²/M) = √(2/5)R

For a ring rotating about an axis perpendicular to its plane, the moment of inertia is Iring = MR2. Therefore, the radius of gyration is:

K_ring = √(MR²/M) = R

Given K_sphere = K_ring, we have:

√(2/5)R_sphere = R_ring

Squaring both sides, we get:

(2/5)R_sphere² = R_ring²

R_sphere² = (5/2)R_ring²

R_sphere = √(5/2)R_ring

Thus, the ratio of the radii is √(5/2). Comparing to the format √(x/2), x = 5. The value is 5, which lies within the range [5,5].