# Sphere rolling in circular bowl

1. Mar 25, 2013

### WannabeNewton

1. The problem statement, all variables and given/known data
This is regarding the following problem: http://s24.postimg.org/bmaa4a65h/sphere_in_bowl.png

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

3. 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 its self 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).

2. Mar 26, 2013

### ehild

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. Mar 26, 2013

### WannabeNewton

Thanks ehild!