Final answer:
The radius of gyration k for the hollow sphere is found by equating the kinetic energies for rolling without slipping and sliding without rolling. By considering the energy conservation and the moment of inertia for a hollow sphere, k is determined to be R/√2, which is option A.
Step-by-step explanation:
When the hollow sphere rolls down an inclined plane without slipping, its kinetic energy at the bottom is a combination of translational and rotational kinetic energy. The translational kinetic energy is (1/2)mv^2 and the rotational kinetic energy is (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity. For rolling without slipping, v = ωR. Thus, Iω^2 can be rewritten as Iv^2/R^2. The moment of inertia for a hollow sphere is I = (2/3)mR^2, so substituting I into the kinetic energy equation gives us a total kinetic energy when rolling of (1/2)mv^2 + (1/3)mv^2 = (5/6)mv^2.
If the sphere slides down the same incline without rolling, there is no rotational kinetic energy, so all the potential energy is converted into translational kinetic energy, which would be (1/2)mv'^2 with v' being the new speed at the bottom. Given that v' is 5v/4, we can say that (1/2)m(5v/4)^2 = (1/2)mgh, where h is the height of the incline. Since both scenarios start with the same potential energy, mgh, we can equate the kinetic energies: (5/6)mv^2 = (1/2)m(5v/4)^2.
Solving for v^2, we see that the (5/6) factor corresponds to the kinetic energy when rolling and the (1/2)*(5/4)^2 factor for when slipping, so (5/6) = (1/2)*(5/4)^2, or (5/6) = (25/32). This simplifies to a ratio of v^2/v'^2 = 6/5. The radius of gyration k can then be found from the ratio of the moments of inertia, where for a hollow sphere, it is k^2 = (2/3)R^2. Equating (1/2)m(6/5)v^2 when rolling down to (1/2)mk^2v^2, we have k^2 = (6/5)(2/3)R^2, which simplifies to k = R/√2, or option A.