What is the total rotational kinetic energy of the wheels?

AI Thread Summary
The discussion centers on calculating the total rotational kinetic energy of a moped's wheels, given a rider's mass of 50kg, the moped's mass of 75kg, and a speed of 20 m/s. The initial calculation for rotational kinetic energy used an incorrect angular velocity, leading to a total of 8 Joules for two wheels. Participants clarify that angular velocity (ω) must be expressed in radians per second, and the relationship between linear speed (v) and angular speed is given by v = rω. To convert degrees to radians, the conversion factor of π/180 is suggested. Accurate calculation requires using the correct angular velocity to determine the true rotational kinetic energy.
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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 rotational 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?
 
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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.)
 
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}
 
thanks. I knew I was forgetting something. It just didn't look right.
 
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