Sphere rolling in circular bowl

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SUMMARY

The discussion focuses on the dynamics of a sphere rolling in a circular bowl, specifically analyzing the motion using the center of mass and the contact point as reference frames. The participant derived the angular frequency of the sphere's motion as ##\omega = \sqrt{\frac{5}{7}\frac{g}{R}}##, confirming consistency with the book's solution that employs a different approach involving friction and torque. The participant questions the necessity of the book's method, suggesting that using the contact point simplifies the problem without significant loss of accuracy. The conversation highlights the importance of choosing appropriate reference frames in solving dynamics problems.

PREREQUISITES
  • Understanding of rotational dynamics and torque equations
  • Familiarity with the parallel axis theorem
  • Knowledge of small angle approximations in physics
  • Basic concepts of angular velocity and motion in circular paths
NEXT STEPS
  • Study the application of the parallel axis theorem in rotational dynamics
  • Explore the implications of the no slipping condition in rolling motion
  • Learn about the derivation and applications of angular frequency in oscillatory systems
  • Investigate different reference frames in dynamics problems for better problem-solving strategies
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Physics students, educators, and anyone interested in understanding the mechanics of rolling motion and the application of rotational dynamics principles.

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).
 
Physics news on Phys.org
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!
 

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