1. Nov 23, 2007

### breez

For a wheel in pure roll, or smooth rolling motion, with constant angular velocity, $$\omega$$, the velocity of the point on the top of the wheel is $$2v_{com}$$, and the velocity of the point at the bottom of the wheel is 0. (all relative to the ground)

However, since centripetal acceleration is $$\frac{v^2}{R}$$, wouldn't that mean the acceleration of the wheel is 0 at the bottom and $$\frac{4v_{com}^2}{R}$$ at the top? This is contradictory to the fact that if $$\omega$$ is constant, $$a_{centripetal}$$ = $$\omega^2r$$, which should be the same throughout the wheel.

2. Nov 23, 2007

### Crosson

But v^2/R applies in the reference frame where the center of the circle is at rest.

3. Nov 23, 2007

### breez

The translational velocity of the wheel should then not affect the centripetal acceleration of the wheel correct? Since with the reference frame of the com of the wheel, would be equivalent to a frame with the wheel at rest. Therefore, the observed acceleration in both cases should still be v^2/R?

4. Nov 23, 2007

### i_island0

to write v^2/R, we should first find R. This R is not the radius of the wheel. The top most point of the wheel follows a cycloidal path. So we should use genral equation to find the radius of the trajectory at the top most point. And that comes out to be 4R... So, acceleration at the top most point comes out to be (2v)^2/4R = v^2/R.
So there is no contradiction again.

5. Nov 23, 2007

### breez

How did you derive 4R as the radius at the top? What is the definition of radius of trajectory?

6. Nov 24, 2007

### Gokul43201

Staff Emeritus
7. Nov 24, 2007

### arildno

The radius R of curvature for a twice differentiable curve, parametrized by $$\vec{r}(t)=(x(t),y(t))$$, is given by the expression:

$$\frac{1}{R}=\frac{||\vec{a}\times\vec{v}||}{||\vec{v}||^{3}}, \vec{v}=\frac{d\vec{r}}{dt},\vec{a}=\frac{d\vec{v}}{dt}$$

Use this combined with the cycloidal representation to derive the correct result for the centripetal acceleration in terms of the radius of curvature.

Last edited: Nov 24, 2007
8. Nov 24, 2007

### Staff: Mentor

As others have explained, if you view the path of a point on the wheel, you can see that its cycloidal path has a radius of 4R at the top (not 2R, or R). But here's another way of viewing things that might prove helpful. What's important to realize is that while the speed of the bottom of the wheel is instantaneously zero with respect to the ground, its acceleration is not.

The top of the wheel is instantaneously in pure rotation about the bottom point. Since its speed with respect to that point is 2v, its centripetal acceleration with respect to that point would be $a = -(2v)^2/(2R) = -2v^2/R$. But that bottom point is itself accelerating (with respect to an inertial frame); its acceleration is $+v^2/R$. So the net acceleration of the top of the wheel with respect to an inertial frame is $-2v^2/R + v^2/R = -v^2/R$, as expected.

9. Nov 24, 2007

### i_island0

how do you say that bottom point of the wheel is accelerating.. According to my calculation, a(radial) is zero as velocity of that point is zero. and a(tangential) is zero as its pure rolling.

10. Nov 24, 2007

### Staff: Mentor

You need to view things from an inertial frame. The center of the wheel is one such frame, since its velocity is constant. All points on the rim of the wheel are centripetally accelerated.

11. Nov 24, 2007

### i_island0

ah..ok.. i agree to it then.. i was seeing that w.r.t ground