Which mass to use in a uniform circular motion? I'm working on a physics question relating to centripetal forces, uniform circular motion, and bike wheels. The tangential velocity and the radius of the bike wheel are known. So to calculate force I know that the equation used is $F=\frac{mv^2}{r}$. But since a bike wheel is a continuous mass, and doesn't have a single identifiable mass like a ball on a string, what is the mass used in the force equation? My hunch is that it is the mass of the entire bike wheel, but I'm not sure and any help would be much appreciated, thanks!
 A: $\newcommand{\vect}[1]{{\bf #1}}$
In this situation you should use the expression
$$
\sum \vect{\tau} = I \vect{\alpha}
$$
where 


*

*$\vect{\tau} = \vect{r}\times \vect{F}$ is the torque

*$I$  is the moment of inertia of the wheel ($I = M R^2$)

*$\vect{\alpha}$ the angular acceleration $\alpha = {\rm d}^2 \theta/{\rm d}t^2$

A: Yes, it is a continuous mass. The Newtonian Mechanics is about point-like masses.
The solution of the problem is that you decompose the continuous body into small (ideally, infinitely small) point-like particles and calculate for them. Calculating with infinitely many, infinitely small mass points requires a little bit of math knowledge (invented partially also by Newton).
The centrifugal force on the points of the bike wheel will sum to zero. Corresponding this, normally the bike wheel remains in place while you are using it (in the frame of the bike). It is because the opposing points of the wheel are accelerated into exactly opposing directions.
You can sum the length of the centripetal force vectors. This sum won't say anything, exactly what force is affecting into which direction.
If you want to calculate some more complex thing (like the force affecting the induvidual spokes, or the force trying to pull the wheel out), then your model is not enough for that, you need a little bit more complex calculations (but not far more).
