Sphere rolling in circular bowl

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The discussion revolves around a physics problem involving a sphere rolling in a circular bowl, focusing on the motion of the sphere's center of mass and the no slipping condition. The solution approach involves using torque equations and the parallel axis theorem to derive the angular acceleration of the sphere. The original poster questions the necessity of using the center of mass as the origin, noting that their method simplifies the problem without significantly increasing complexity. The conversation highlights the validity of different approaches to solving the problem, emphasizing that both methods yield the same result. The discussion concludes with a recognition of the complexities involved in choosing an appropriate coordinate system for analysis.
WannabeNewton
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Homework Statement


This is regarding the following problem: http://s24.postimg.org/bmaa4a65h/sphere_in_bowl.png

Homework Equations


I will be referring to this drawing: http://s22.postimg.org/6zl9mw9zj/drawin.png

The Attempt at a Solution


Let me just show first how I got my answer and I'll then state my question. First note that the center of mass of the sphere travels in a circular trajectory of radius ##R - b## hence ##v_{cm} = (R - b)\dot{\varphi}##. The constraint here is the no slipping condition which tells us that, in this case, the velocity of the contact point at the given instant should be zero i.e. ##(R - b)\dot{\varphi} = b\dot{\theta}## where ##\dot{\theta}## is the angular velocity of the sphere due to itself rotation about its center of mass. Fixing the origin to the contact point, for the given instant, the torque equation becomes ##(\frac{2}{5}mb^{2} + mb^{2})\ddot{\theta} = \frac{7}{5}mb^2\ddot{\theta} = -mgb\sin\varphi## where I have used the parallel axis theorem. Using the constraint, this reduces to ##\frac{7}{5}b\ddot{\varphi}(R - b) = -gb\sin\varphi##, which after applying the small angle approximation gives us ##\ddot{\varphi} + \frac{5}{7}\frac{g}{R(1 - \frac{b}{R})}\varphi = 0##. Using the fact that ##R >> b## this gives us ##\omega = \sqrt{\frac{5}{7}\frac{g}{R}}##.

Now I checked the book's solution and it gives the exact same answer but it instead chooses the center of mass of the sphere as the origin and looks at the torque due to friction and then uses Newton's 2nd law to solve for friction etc. etc. Is all of this even necessary? Granted it doesn't seem to make the work MUCH longer but what's the point? The problem itself is simple if you just take the instantaneous origin to be on the contact point (assuming what I did was actually correct lol).
 
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The torque equation Idω/dt=τ is valid for a fixed axis or with respect to the CM.
The contact point is fixed instantaneously, but the coordinate system attached to it is accelerating. You can find the instantaneous acceleration, but it is not convenient to describe the motion.

ehild
 
Thanks ehild!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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