What Is the Correct Kinetic Energy of Rotating Tires?

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SUMMARY

The correct kinetic energy of rotation for four tires of an automobile traveling at 32 m/s with a radius of 22 cm and mass of 11 kg is calculated using the formula KE = 1/2 Iw^2. The moment of inertia (I) for a disk is 0.5 * m * r^2, which results in I = 0.5 * 11 kg * (0.22 m)^2 = 0.5324 kg·m². The angular velocity (w) is determined by w = v/R, yielding w = 145.45 rad/s. The total kinetic energy for all four tires is KE_total = 4 * (1/2 * 0.5324 * (145.45)^2) = 5630 Joules.

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Homework Statement



An automobile is traveling at v = 32 m/s and has wheels with radius R = 22 cm. Assuming that the wheels are disks which roll without slipping and have a mass m = 11 kg, what is the kinetic energy of rotation of the four tires in Joules?


Homework Equations



w = v/R

The kinetic energy for each tire is then


KE = 1/2 Iw^2

There are four wheels.

The Attempt at a Solution



I tried converting 32 m/s to rad/s for angular velocity. So 32m/s divided by .22m (radius) to get 145.45.


Then I squared that answer. I multiplied it by .5.

Then for I, I used mr^2 and got .5324.

I got 5631.648. Tried multiplying that by four and it didn't work.

I've tried entering

3 Incorrect. (Try 1) 5630
5 Incorrect. (Try 2) 5.6x10^2
7 Incorrect. (Try 3) 2.3x10^4
9 Incorrect. (Try 4) 22500


None of these were correct. I'm wondering where I messed up...
 
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"Then for I, I used mr^2 and got .5324."

The wheels were assumed disks, and the moment of inertia of a disk is 0.5 mr^2.

ehild
 

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