# What is the total rotational kinetic energy of the wheels?

1. Oct 20, 2006

### brncsfns5621

A 50kg rider on a moped of mass 75kg is traveling with a speed of 20 m/s. Each of the two wheels of the moped has a radius of 0.2m and a moment of inertia of 0.2 kg*m^2. What is the total rotaional kinetic energy of teh wheels?

KEr = 1/2Iw^2
= 1/2(.02)(20)^2
= 4
Two wheels = 2 *4 = 8 Joules

I think I am wrong as the angular velocity (w) has to be in rad/s. For some reason, I can not figure out how to get that. What is the formula for w?

2. Oct 20, 2006

### Staff: Mentor

Yes, that formula requires angular speed, not linear speed. They are related by: v = wr.

(A complete circle, one circumference, is equivalent to $2 \pi$ radians.)

3. Oct 20, 2006

### NateTG

Yeah, you're using an incorrect value for $\omega$.

If you're confused by the radians, it might be helpful to figure out how many degrees per second the wheels are turning, and then use the conversion:
$$1 \rm{degree}=\frac{\pi}{180} \rm{radians}$$

4. Oct 20, 2006

### brncsfns5621

thanks. I knew I was forgetting something. It just didn't look right.