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

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SUMMARY

Two spherical balls of different sizes and weights roll down a hill without slipping at the same speed due to the conservation of mechanical energy. The equation E = mgh + (1/2)m*v^2 + (1/2)I*ω^2 governs their motion, where I represents the rotational inertia. Despite initial assumptions that size affects speed, the relationship between mass, radius, and rotational inertia leads to equal speeds for both balls. The key insight is that the rotational inertia I is proportional to the mass and the square of the radius, balancing the energy distribution between translational and rotational kinetic energy.

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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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