How to Calculate Total Kinetic Energy of a Car with Rotational Motion?

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To calculate the total kinetic energy of a car with rotational motion, both translational and rotational kinetic energy must be considered. The formula for total kinetic energy includes the translational component (1/2 mv^2) and the rotational component (1/2 Iω^2), where I is the moment of inertia and ω is the angular velocity. The discussion highlights the importance of including the rotational inertia of the tires and wheels to avoid significant errors in calculations. For part (c), the acceleration can be determined using F=ma, but neglecting rotational inertia leads to inaccuracies in results. Understanding these concepts is crucial for accurately solving the problem and calculating the fraction of kinetic energy in the tires and wheels.
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Homework Statement


The 1900 kg mass of a car includes four tires, each of mass (including wheels) 40 kg and diameter 0.80 m. Assume each tire and wheel combination acts as a solid cylinder.
(a) Determine the total kinetic energy of the car when traveling 60 km/h.


(b) Determine the fraction of the kinetic energy in the tires and wheels.
%

(c) If the car is initially at rest and is then pulled by a tow truck with a force of 2500 N, what is the acceleration of the car? Ignore frictional losses.


(d) What percent error would you make in part (c) if you ignored the rotational inertia of the tires and wheels?


Homework Equations


I would understand where to go in step two. For step 1 i was using K=1/2 mv^2
For step 3 i used F=ma. Apparantly I am totally wrong.

The Attempt at a Solution


Help. The formulas I use are giving me incorrect results. about 200,000 J is what I got
 
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You're forgetting the rotational kinetic energy stored in the rotation of the wheels. In addition to the \frac{mv^2}{2} there should be a \frac{I\omega^2}{2} term.
 
the rotational kinetic energy can be found from
I=1/2*40*4*.4^2?
and
60*1000/3600=v
omega=v/0.4?
 
check that. I was able to solve part a)..thankyou!
I do not even know how to start on the next 3 parts..if you could point me where to go I would be indebted.

wouldnt b simply be to divide the rotational Kinetic energy by the total kinetic energy? apparently not..
 
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