How Is the Total Kinetic Energy of a Car Calculated?

AI Thread Summary
The total kinetic energy of a car and its wheels can be calculated using the formula K_tot = 4(0.5I_CMω² + 0.5MV_CM²) + K_car. In this case, the wheels are modeled as homogeneous cylinders with a weight of 25 kg each and a radius of 0.3 m, while the car itself weighs 1000 kg. The initial calculation mistakenly omitted squaring the speed in the last term, leading to an incorrect total kinetic energy of 82.2 kJ instead of the correct 518 kJ. The correct approach involves ensuring all terms are properly squared and accounted for. Accurate calculations are crucial for determining the total kinetic energy in physics problems.
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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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