How Is the Total Kinetic Energy of a Car Calculated?

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SUMMARY

The total kinetic energy of a car and its wheels moving at a speed of 30 m/s is calculated using the formula K_{tot}=4(0.5I_{CM}\omega^{2}+0.5MV_{CM}^{2})+K_{car}. The wheels, modeled as homogeneous cylinders, each weigh 25 kg with a radius of 0.3 m, while the car itself weighs 1000 kg. The correct total kinetic energy is 518 kJ, which was miscalculated in the discussion due to an error in squaring the speed in the last term of the equation.

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fishingspree2
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Homework Statement


The 4 wheels of a car each weigh 25kg and have a radius of 0,3m. Without its wheels, the car weighs 10^3 kg. The wheels are homogenous cylinders. What is the total kinetic energy of the car and the wheels if the car is moving at a speed of 30 m/s?


Homework Equations


K_{tot}=4(0.5I_{CM}\omega^{2}+0.5MV_{CM}^{2})+K_{car}
the first to terms of the last question concern the wheels
I_{cylinder}=\frac{MR^{2}}{2}
\omega=\frac{V_{CM}}{R}

The Attempt at a Solution


replacing omega by V/R, I by MR^2/2 and distributing the 4:
K_{tot}=MV^{2}+2MV^{2}+\frac{10^{3}\times30}{2}
K_{tot}=25\times30^{2}+50\times30^{2}+\frac{10^{3}\times30}{2}=82,2 \mathrm{kJ}
The right answer is 518 kJ

where is the mistake? :(
thank you, sorry for my english
 
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fishingspree2 said:

The Attempt at a Solution


replacing omega by V/R, I by MR^2/2 and distributing the 4:
K_{tot}=MV^{2}+2MV^{2}+\frac{10^{3}\times30}{2}

You forgot to square the speed in that last term.
 
Thank you very much
sorry for the waste
 

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