Ball Rotation on a Ball: Solving for Center of Mass Velocity

[Nathanael]

Also, could someone explicitly state what is meant by theta with a dot on top? I don't recall this notation.

$\dot \theta$ is shorthand notation for the time derivative of θ (that is, $\dot \theta \equiv \frac{d\theta}{dt}$)

More generally $\dot x$ is defined to mean $\frac{dx}{dt}$ where x can mean pretty much anything, but t exclusively means time. (In other words, dy/dx would not be written as $\dot y$ unless x represents time.) I think this notation is used primarily in physics, but it is used quite a bit.

Furthermore, $\ddot x\equiv \frac{d^2\theta}{dt^2}$

 

[Karol]

Nathanel already answerd, i erased mine

 

[Nathanael]

Karol, you can save time/effort by using the "constraint of rolling" which is $v_{cm}=R\dot \phi$. That solves the problem without needing to consider $\dot \theta$

I think Haruspex (when talking about tangent ramps) was just trying to show you that this equation is just as true on a sphere as it is on a flat surface.
(The ball rolls dΦ, it's center of mass must move a distance RdΦ, and everything is flat in this differential limit, thus $v_{cm}=R\dot \phi$)

 

[haruspex]

Karol, you can save time/effort by using the "constraint of rolling" which is $v_{cm}=R\dot \phi$. That solves the problem without needing to consider $\dot \theta$

I think Haruspex (when talking about tangent ramps) was just trying to show you that this equation is just as true on a sphere as it is on a flat surface.
(The ball rolls dΦ, it's center of mass must move a distance RdΦ, and everything is flat in this differential limit, thus $v_{cm}=R\dot \phi$)

Not exactly.
The ball's mass centre is moving at $(R+r)\dot \theta$, and it is rolling on a stationary surface, so its rotation rate is $(R+r)\dot \theta/r$. From that you can write down the KE straight away. I mentioned a ramp because I thought it make that viewpoint more apparent.

 

[Karol]

Thanks all