hollow vs. solid sphere - rotational inertia

[dadasolee]

Why is it that the translational speed of a hollow sphere as it reaches the bottom of an inclined plane is higher than that of a solid sphere? Perhaps because it reaches the bottom later than the solid sphere, which allows longer time to gain speed?

Fnet = mgsin0-Ia/R^2
Rotational Inertia of hollow sphere: (2/3)mR^2
Rotational Inertia of solid sphere: (2/5)mR^2

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[thebillsfan]

i wouldve thought its the other way around...if the center of mass is closer to the center of the sphere, in order to maintain angular momentum, it would have to speed up (think ice skater)??

 

[loveoforganic]

i wouldve thought its the other way around...if the center of mass is closer to the center of the sphere, in order to maintain angular momentum, it would have to speed up (think ice skater)??

ditto

 

[wanderer]

i wouldve thought its the other way around...if the center of mass is closer to the center of the sphere, in order to maintain angular momentum, it would have to speed up (think ice skater)??

Yea more work has to be done on the object with a greater I to achieve the same linear velocity, much like more work has to be done on something with greater inertia/mass to get to the same vf, so the solid sphere will be moving faster at the bottom of the plane.

 

[iPodtosis]

Yea more work has to be done on the object with a greater I to achieve the same linear velocity, much like more work has to be done on something with greater inertia/mass to get to the same vf, so the solid sphere will be moving faster at the bottom of the plane.

You mean hollow sphere move faster at the bottom right? cause hollow sphere has greater Inertia, need more work, more kinetic energy, more speed.

 

[wanderer]

You mean hollow sphere move faster at the bottom right? cause hollow sphere has greater Inertia, need more work, more kinetic energy, more speed.

Okay
(X is the fraction in front of the Mr²)
so mgh = .5mv² + Iw²
mgh = .5mv² + I * (v²/r²)
mgh = .5mv² + X Mr² * (v²/r²)
gh = .5v² + X r² * (v²/r²)
gh = v²(.5+X)
sqrt (gh / (.5+X) ) = v
..........................

 

[Seraph 84]

A hollow sphere will have a much higher moment of inertia I. Since it's rolling down an incline, we can apply conservation of mechanical energy to the sphere, where KE = PE. Now, since it has a moment of inertia, not all of the PE will be converted directly into translational kinetic energy - some of it is converted into rotational kinetic energy.

KE = 1/2mv^2 + 1/2Iw^2

See, if the sphere is hollow, it will have more rotational inertia, and more of that energy will be used to keep the ball rolling than translating. If it were a box, then we could eliminate the second term in the equation and the translational energy would be equal to the kinetic energy. This concept explains why boxes will reach the end of an incline before any object that rolls.

 

[iPodtosis]

but according to OP
Rotational Inertia of solid sphere: (2/5)mR^2
Rotational Inertia of hollow sphere: (2/3)mR^2
so solid has less inertia than hollow

 

[Seraph 84]

Hmm, you're right it is, I thought it was the other way around. Solid sphere should reach the bottom of the incline first then.

 

[dadasolee]

Thanks for all the discussions🙂 My mistake, it was the other way around, and yes, the hollow sphere due to its greater rotational inertia would reach the bottom later with lower translational speed.

 

[Doodl3s]

Is this even ON the MCAT? its not in Kaplan's or EK's book on physics