Final answer:
For a solid sphere rolling without slipping down an incline, the fraction of kinetic energy that is rotational is 2/7. This is determined by comparing the gravitational potential energy at the top with both the translational and rotational kinetic energy at the bottom of the incline, using the moment of inertia for a solid sphere.
Step-by-step explanation:
The question is asking about the kinetic energy distribution for a solid sphere rolling down an incline without slipping. Kinetic energy in this scenario is of two types: translational and rotational. Since the sphere is rolling without slipping, its translational kinetic energy (KE₁) is tied to its rotational kinetic energy (KE₂) through the relationship KE₁ = KE₂.
For a solid sphere, this is due to the fact that the moment of inertia (I) is (2/5)mr² and the relationship between rotational kinetic energy and moment of inertia is KE₂ = (1/2)Iω², where ω is the angular velocity. When the sphere rolls down the incline, gravitational potential energy (PE₀ = mgh) is converted into both forms of kinetic energy, giving us the equation PE₀ = KE₁ + KE₂.
As the sphere rolls down the incline, it gains angular velocity due to the rotational acceleration caused by the torque from the gravitational force component along the incline. The fraction of kinetic energy that is rotational for a sphere is essentially the ratio of rotational kinetic energy to the total kinetic energy at any point during its motion down the incline.
So, if we equate the gravitational potential energy at the top of the incline to the sum of translational and rotational kinetic energies at the bottom of the incline, using the relationships for a solid sphere (I = (2/5)mr² and ω = v/r), we get PE₀ = (1/2)mv² + (1/2)*(2/5)mr²*(v/r)²=> mgh = (7/10)mv². The rotational kinetic energy is then (2/5)mv², hence the fraction of the total kinetic energy that is rotational is 2/7.