Final answer:
To find the moment of inertia of a rigid cylindrical body rolling down an incline, the principles of energy conservation and rotational dynamics are used. The process involves equating the body's total kinetic energy at the bottom to its initial potential energy and solving for a moment of inertia, leading to the answer: b) I = 0.5mr².
Step-by-step explanation:
The problem involves determining the moment of inertia of a rigid body with a cylindrical cross-section rolling down an incline. We can use the principles of energy conservation and the fact that the body rolls without slipping to find the solution. The kinetic energy of the rolling object is the sum of its translational and rotational kinetic energies. At the top of the incline, the body has potential energy, which is entirely converted to kinetic energy as the body reaches the bottom.
Using the equation for kinetic energy (K = ½mv² for translational and K = ½Iω² for rotational), and the fact that v = rω for rolling without slipping, we can equate the total kinetic energy to the potential energy (mgh), where h is the vertical height of the incline which can be found using the sine function (sin(30°) × 10.0 m). By combining these equations and solving for the moment of inertia I, we can express it as a multiple of mr². After performing the calculations, we find that the correct option is b) I = 0.5mr².