Why Do Spherical Balls of Different Sizes Roll at the Same Speed Downhill?

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Both spherical balls, regardless of size or weight, roll down a hill at the same speed due to the conservation of mechanical energy. The gravitational potential energy converts into both translational and rotational kinetic energy, but the ratio of these energies remains consistent across different sizes. The moment of inertia (I) for uniformly dense spheres relates to their mass (m) and radius (R), influencing their rotational motion. The relationship between linear velocity (v) and angular velocity (ω) also plays a crucial role in ensuring that both balls reach the bottom simultaneously. Understanding these principles clarifies why size does not affect the rolling speed in this scenario.
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Homework Statement


Suppose two spherical balls roll down a hill without slipping. If both are released from rest, which one will roll the fastest?

The answer is that both will roll at the same speed, even if they are of different sizes and weights, but I do not understand why

Homework Equations


Conservation of mechanical energy E= mgh + (1/2)m*v^2 + (1/2)I*ω^2

The Attempt at a Solution


I would think that the smallest ball rolls the fastest. If all of the gravitational potential energy mgh is turned into kinetic energy (1/2)m*v^2 + (1/2)I*ω^2, then the bigger ball would lose more energy to rotational kinetic energy because the rotational inertia I is greater because the radius is bigger. However, that is not the correct answer. I think that friction is not considered in this problem.
 
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dhphysics said:

Homework Statement


Suppose two spherical balls roll down a hill without slipping. If both are released from rest, which one will roll the fastest?

The answer is that both will roll at the same speed, even if they are of different sizes and weights, but I do not understand why

Homework Equations


Conservation of mechanical energy E= mgh + (1/2)m*v^2 + (1/2)I*ω^2

The Attempt at a Solution


I would think that the smallest ball rolls the fastest. If all of the gravitational potential energy mgh is turned into kinetic energy (1/2)m*v^2 + (1/2)I*ω^2, then the bigger ball would lose more energy to rotational kinetic energy because the rotational inertia I is greater because the radius is bigger. However, that is not the correct answer. I think that friction is not considered in this problem.

You aren't thinking very hard about this. Assume they are all uniformly dense spheres. Now find a relation between I and the mass m and R and a relation between ω and v and R. Now what do you think?
 
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