enter image description here

These are how circular turning tracks look like. Now , according to my textbook:enter image description here

This is how it looks like. But I argue upon each & every point of it.

  1. Here , the car is turning but it’s direction of WHEELS is straight , which implies that direction of velocity of car is tangentially straight. How is that possible ? If the car is turning , then the front wheels of car also turn & are not straight in direction. Only the direction of back side wheels of car is straight.

Even if I think for a time dt covered by car , then also car is going to turn in direction which I think is VT but not straight. Where am I going wrong ?

EDIT: I was revising Acceleration in circular motion & I got to know that a particle moving in a circle has 2 components. Radial & tangential vectors. We know if there are two component vector , then they have to be of s Resultant vector i.e vector in between the two vectors. So , I think the V in the diagram above has to be tangential velocity & centripetal velocity is just not shown in the diagram. Am I right ? enter image description here

  • 1
    $\begingroup$ Perhaps obviously, in a turning car the direction of the wheels is not straight. There's also the differential that lets the wheels on both sides turn at different speeds. But those are needless complications for your question. Location matters. The tangential velocity is for the car is assigned to the center of mass of the car, not between the two front wheels. Instead of thinking about a car, think about a unicycle or a Segway. Your title doesn't match your question, by the way. $\endgroup$
    – DKNguyen
    Dec 24, 2021 at 18:42
  • $\begingroup$ In a circular motion, the linear velocity is "conceptually" the equivalent of the rotational speed on a piece-wise straight line that is tangent to the curve. It is similar to the resulting shear force due to torque, which is indicated by a straight line, but the object is curving. $\endgroup$
    – r13
    Dec 24, 2021 at 21:15
  • $\begingroup$ If you look closely, the diagram shows the velocity angled in a bit. If you take the CG location of the vehicle, the CG velocity vector will be tangential to the instantaneous path of the CG, which is perpendicular to the instantaneous center of motion of the CG. $\endgroup$
    – Phil Sweet
    Dec 26, 2021 at 20:37

2 Answers 2


The car steering is designed to set the alignment of the four wheels to follow the curve of rotation without any skidding.

At any $\Delta t$ time the car experiences two components of force, tangential$F_t$ and centripetal $F_{centripetal}=mv^2/r$

If the car is not accelerating tangentially, the $F_t=0$ and the only force acting on the car is $F_s,$ which is equal and opposite of the centripetal force as your diagram.



  • $\begingroup$ If Ft = 0 & only centripetal is there. Then , the car would just move towards centre. It wouldn’t even turn. Why ? Because the total resultant force is only towards centre axis & there is no force forward. 2) A car stays in motion until a constant force is applied on it. I’m really confused. $\endgroup$
    – S.M.T
    Dec 25, 2021 at 9:43
  • $\begingroup$ I do agree impulse is present in forward direction but if there is no Ft, then how can it keep on moving forward. Also , if I consider the case of impulse. I can say there are two types of impulse present. One forward & other radial. $\endgroup$
    – S.M.T
    Dec 25, 2021 at 10:00

If you are implying that the drawing does not show the wheels turning its just because this is a simplification.

The velocity is always tangential to the trajectory in space of the material point.

However, because there is a (constant) force ($f_s$) towards the inside of the circle what applies here is the law of impulses. I.e.

$$\vec{v}_i + \vec{f}_S dt = \vec{v}_{f}$$

Notice the vector notation. The initial velocity is tangential, but the force $f_s$ is normal to the tangent (i.e. along the radius). Therefore what happens is that indeed the summation of the initial velocity and the impulse of $f_s$ result in what you have drawn as $v_T$


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