# Rolling motion

1. Mar 25, 2005

### klandestine

I know that because of the rolling constraint, V(cm)=R*w [w=omega]. Also, if I want to find the velocity of particle i, i can sum V(cm) and V(i, relative to the center of mass). But how do I determine V(i, rel)? For example, if I am given the speed and radius of a wheel, how fast is a point at the very front edge of the wheel going? I keep trying to use these equations, but I keep coming up with 2wR, which I know is the velocity for the TOP of the wheel.

2. Mar 25, 2005

### Staff: Mentor

Realize that the velocity of a particle relative to the cm is a vector whose magnitude is given by $\omega R$ and whose direction is tangent to the wheel. So, if the velocity of the cm is $V \hat{x}$, then the velocity of a point at the front edge relative to the cm is $-V \hat{y}$. The velocity relative to the ground of that point is $V \hat{x} -V\hat{y}$, giving a speed of $\sqrt{2} V = \sqrt{2} \omega R$.

3. Mar 25, 2005

### klandestine

ah-ha!

thank you! so, if i wanted to know the velocity of a point at a more arbitrary location, i would have to use trigonometry, right?

4. Mar 25, 2005

### Staff: Mentor

Right. You'd have to add the two velocities as vectors, which will involve a bit of trig.