Why Does Sphere's Angular Velocity Differ with Different Reference Points?

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SUMMARY

The discussion centers on calculating the angular velocity of a uniform solid sphere rolling inside a hemispherical bowl. The correct formula for angular speed when the sphere reaches the bottom is derived as ω² = [10gR(1-cos θ)]/7r², where R is the radius of the bowl, r is the radius of the sphere, and θ is the angle to the vertical. The confusion arises from using different heights for potential energy calculations, but the solution using the height of the sphere's contact point with the bowl is validated as correct. The reasoning emphasizes the importance of reference points in determining angular velocity.

PREREQUISITES
  • Understanding of rotational dynamics and angular velocity
  • Familiarity with the concepts of potential and kinetic energy
  • Knowledge of the moment of inertia for a uniform solid sphere (I = 2mr²/5)
  • Basic trigonometry to analyze angles and height in the context of physics problems
NEXT STEPS
  • Study the principles of rolling motion and energy conservation in physics
  • Learn about the moment of inertia for various shapes and its applications
  • Explore the effects of different reference points on angular motion
  • Investigate advanced problems involving rolling objects in varying geometries
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Students of physics, particularly those studying mechanics, educators teaching rotational dynamics, and anyone interested in the application of energy conservation in rolling motion scenarios.

Dorothy Weglend
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Homework Statement


A uniform solid sphere of radius r is placed on the inside surface of a hemispherical bowl with much larger radius R. The sphere is released from rest at an angle theta to the vertical and roles without slipping. Determine the angular speed of the sphere when it reaches the bottom of the bowl.


Homework Equations


Kinetic energy of rolling motion = Iw^2/2 + mw^2r^2/2
Potential energy of sphere = mgh.
I=2mr^2/5 for uniform sphere.

The Attempt at a Solution



I am not sure what to use for h. At first I thought that I should use the height of the center of mass of the sphere, so h = (R-r)(1-cos(theta)). Doing all the substitutions and solving for omega:

w^2 = [10g(R-r)(1-cos theta)]/7r^2

Then I thought that since the angular velocity should be the same at all points of the disk, I can use the height of the bottom of the sphere, h=R(1-cos theta):

w^2 = [10gR(1-cos theta)]/7r^2

Then I started wondering, why, if the angular speed is the same, do I get two different answers for it?

Then I realized that this makes sense, since the point on the rim of the sphere would be rotating around the axis of the sphere as well as around the center of the bowl. That would make the angular speed greater, as we see here. So, obviously, the first answer is the one to use, since the CM is not rotating, just moving parallel to the surface of the bowl, and so its angular velocity around the center of the bowl would give the right speed.

So, very pleased with my beautiful line of reasoning, I checked the answer, only to discover that I am wrong, since the given answer is

w^2 = [10gR(1-cos theta)]/7r^2

which is the same as my second answer, using the point of the ball resting on the surface of the bowl, not the CM.

However, I can't come up with the reason why this is the better solution :confused:

Can someone help me understand this, please?

Thank you,
Dorothy
 
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Obviously, your first solution is the correct one. One can see that by looking at a pathological limiting case: Imagine R ~ r. In that case, the ball almost won't move (a very small rotation of the sphere will displace strongly the point of contact).

If I have to justify the answer in the book, then I'd say that because they stated that R>>r, we can replace R-r by R without much problems. But it is a meager excuse.
 
Thank you, Vanesch. I appreciate your help.


Dorothy
 

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