Sphere rolling in circular bowl

In summary, the student attempted to find the center of mass of a sphere and solved for the torque due to friction.
  • #1
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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  • #2
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
 
  • #3
Thanks ehild!
 

1. How does the shape of the bowl affect the speed of the rolling sphere?

The shape of the bowl can greatly affect the speed of the rolling sphere. If the bowl is shallow and wide, the sphere will roll more slowly due to the smaller angle of incline. If the bowl is deeper and narrower, the sphere will roll faster due to the steeper angle of incline.

2. Does the weight of the sphere impact its rolling motion in a circular bowl?

Yes, the weight of the sphere can impact its rolling motion in a circular bowl. A heavier sphere will have more momentum and will roll faster, while a lighter sphere will have less momentum and will roll slower.

3. How does the friction between the bowl and the sphere affect its movement?

The friction between the bowl and the sphere can greatly impact its movement. If there is a lot of friction, the sphere will roll more slowly and may even stop before reaching the bottom of the bowl. If there is less friction, the sphere will roll faster and may even continue to roll after reaching the bottom of the bowl.

4. Can the shape of the sphere affect its rolling motion in a circular bowl?

Yes, the shape of the sphere can impact its rolling motion in a circular bowl. A perfectly round sphere will roll more smoothly and consistently, while a sphere with an irregular shape may have a more unpredictable and uneven rolling motion.

5. How does the angle of the bowl affect the path of the rolling sphere?

The angle of the bowl can greatly impact the path of the rolling sphere. A bowl with a steeper angle will cause the sphere to roll in a more direct and downward path, while a bowl with a shallower angle may cause the sphere to roll in a more curved and sideways path.

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