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A car has a mass of 2Mg and a center of mass at G . Determine the acceleration of the car , if the rear "driving" wheels are always slipping , whereas the front wheels are free to rotate . Neglect the mass of the wheels . The coefficient of kinetic friction between the wheels and the road is 0.25 .

I want "only" to check if I understood the question.

(the rear "driving" wheels are always slipping , whereas the front wheels are free to rotate) Does it mean that the rear wheels are connected to the engine ,while the front wheels are not ?

How will neglecting the mass of the wheels simplify the problem ? My expectation : (Maybe neglecting the mass of the wheels means "mass moment of inertia*angular acceleration =0 , so rotation is neglected i.e treat the wheels as if they are translating , no rolling.)

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    $\begingroup$ You are correct in both. Wheels free to rotate are not "driven" by the engine - you could say not connected. Neglecting the mass of the wheels simplies the situation because you don't have to take the rotational inertia into account. $\endgroup$
    – Steeven
    Commented Jul 22, 2017 at 13:23
  • $\begingroup$ @Steeven Is there any other simplification resulting from neglecting the mass of the wheels ? $\endgroup$
    – Eman.suradi
    Commented Jul 22, 2017 at 13:32
  • $\begingroup$ On the basis of the phrasing of the question, no, I think Steeven has identified the only important point regarding the wheels. $\endgroup$
    – user163104
    Commented Jul 22, 2017 at 14:43

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If the wheels were massive, they would have a non-negligible moment of inertia. In order to solve the problem, you would need to calculate the torques on each of the wheels from friction in order to get the wheels spinning. Neglecting the wheels means that all the force from friction does work towards the car's translational motion.

The front wheels being free to rotate implies that they are not being driven by the engine. This means they can roll with the road instead of slipping extremely fast like the back wheels. What this does for you is that it greatly reduces the amount of friction produced by the front wheel. When rolling, the only friction present is whatever static friction is enough to keep it from slipping, not the full $\mu_s|\overrightarrow N|$. And since you're ignoring the wheels' masses, this friction force can be assumed to be 0.

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  • $\begingroup$ "Neglecting the wheels means that all the force from friction does work towards the car's translational motion." I didn't understand this sentence Sir $\endgroup$
    – Eman.suradi
    Commented Jul 22, 2017 at 17:55
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    $\begingroup$ Rotational kinetic energy also exists. When an object spins, it has kinetic energy. When an object moves from one location to another, it has kinetic energy. The wheels spinning hold kinetic energy in addition to the translational kinetic energy of the car moving. If the wheels are massless, they have no moment of inertia, and therefore can' t have rotational kinetic energy. $\endgroup$
    – Johnathan Gross
    Commented Jul 22, 2017 at 18:02
  • $\begingroup$ I want to check that I understand your answer . Since the back wheels are slipping very fast , this means the distance travelled by the wheels will be less , so the angular acceleration of the front wheels will be small (since the distance travelled is small) , and thus the friction produced by the front wheels is small , true ? $\endgroup$
    – Eman.suradi
    Commented Jul 22, 2017 at 18:14
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    $\begingroup$ No. Since the back wheels are slipping, only static friction can act. The front wheels aren't being powered by the engine, so they aren't being forced to move. Because they aren't being forced to move, they can stay in contact with the ground without slipping. No slipping means static friction, which can be any value between 0 and $\mu_sN$. Keeping the wheel rolling instead of slipping doesn't require as much force as pushing the whole car. $\endgroup$
    – Johnathan Gross
    Commented Jul 22, 2017 at 19:33
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    $\begingroup$ The back wheels are slipping so the bottom is moving backwards. Friction opposes the backwards motion, by pushing forwards. $\endgroup$
    – Johnathan Gross
    Commented Jul 22, 2017 at 20:04

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