We consider a wheel with radius r that rolls without slipping. This means that the distance d is equal to arc-length **s = rθ,**

**d = rθ** ….. …… (1)

If the center of the wheel travels the distance d in the time t, we have for the linear speed v = d/t , and for the angular speed to **ω = θ/t** . We find **v = (rθ)/t = r(θ/t)** or

**V _{t} = rω** …. …. (2)

Similarly, if the center of the wheel accelerates from v_{0} = 0 to the velocity v in time t, the angular velocity increases from ω_{0} to ω . Since **v = rω** , we find

**A _{t} = rα** … …… (3)

Here, v and a are the velocity and acceleration of a point on the perimeter relative to the the center: i.e., they are the tangential velocity and acceleration, **v _{T} = ωr** and

**Example:** a) A car travels at the speed v = 25 m/s. Find the angular speed of the wheels when the radius of the wheel is r = 0.31 m (1 foot). b) If the car accelerates from rest to v = 25 mis in the time t = 17s, find the angular acceleration of the wheel, and the number of turns of the wheeL

**Solution:** We find for the angular speed,

**ω = v/t = 25m / s = 80.6 (rad/s)**

We find for the angular acceleration,

**α = (ω – ω _{0})/t = (80.6 (rad/s) – 0) / 17s = 4.74 (rad/s^{2})**

The speed is constant for uniform circular motion speed. Since the angular acceleration is zero α = 0, and the tangential acceleration is zero as well. Since v = ωr, we find for the centriptal acceleration towards the center, **a _{c} = v^{2}_{t}/r** or,

**A _{c} = vω^{2}.**

For rolling motion of a wheel without slipping, the linear speed of the center must be the same as tangential speed: **v _{CM} = rω**.