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While turning on a road, why does a car slip towards outside, if we are observing the car from inertial frame of reference, i.e center?

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  • $\begingroup$ Center of the circle $\endgroup$ – Ayush Singh Jun 20 at 7:21
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If you are viewing the car from an inertial frame of reference, such as someone sitting on the side of the road, the car "slips outside" because the car wants to go in a straight line. To follow the turn (instead of going straight), something must apply a force to the car. In this case, it's the wheels of the car pushing on the road which applies the force which causes the car to go in a not-straight path.

If you view this from inside the car, it is not an inertial frame of reference. During the turn, the car is accelerating (acceleration for these purposes is defined as a changing velocity, not just a changing speed. Changing directions counts as accelerating). Thus you will have effects such as centripetal acceleration.

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For any object to go in a circular path, an external force must act on it.

[Inertial frame of reference] Consider a car of mass $m$ moving in a circular path of radius $R$ with coefficient of friction $\mu$. In this case, the frictional force $\mu mg$ acts as a centripetal force that causes the car to move in a circular path.

If the force of friction is less than the centripetal force needed for that car to turn $\left(F_c = \frac{mv^2}{R}\right)$, the car would slip and will turn in a larger radius $\left(\because F_c \propto \frac{1}{R}\right)$.

Note: Centripetal force is not a new kind of force. Any force that causes an object to move in a circular path acts towards the center of that circular path and is called centripetal force.

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I think the much better understanding would be: When we turn on road while moving in a straight line.The inertia tends to keep us in that position.And as a result a pseudo force arises in non inertial frame.But in our frame its just inertia.

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