# Why, when we calculate forces in banking of roads, we do not take the centripetal force but the centrifugal force?

I know the we only apply centrifugal force when in non inertial frame of the object but in banking of road questions, why cant we make equations from inertial frame rather than non inertial frame?

Above are the equations for a car in the frame of the car on questions of a banked road. Here we take the centrifugal force $$m\frac{v^2}{r}$$

But cant we make equations from the inertial frame(ground) and take $$m\frac{v^2}{r}$$ to the right rather than left. If yes then how?

• Just depends on which frame you want to do calculations in… Commented Jun 18 at 19:37
• There is no reason to use centrifugal anything ever, in introductory mechanics. It is just bad teaching. You can, and should, just use centripetal everywhere, consistently. Commented Jun 19 at 1:22
• Commented Jun 19 at 5:21

Define the "outward" direction in your diagram (to the left) to be the positive $$x$$-direction, and define the vertical direction to be the $$y$$-direction.
In the inertial reference frame, Newton's Second Law says that if the car is executing uniform circular motion in a horizontal plane, then the forces acting on the car must add up to be a net inwards force of magnitude $$m v^2/r = m r \omega^2$$. Thus, we will have $$\sum F_x = -m r \omega^2; \qquad \sum F_y = 0.$$ where the sums run over all of the "true" forces acting on the car.
If, on the other hand, we look at the motion in a frame rotating with angular velocity $$\omega$$, then there is an additional centrifugal force of $$m r \omega^2$$. However, the car is at rest in this frame, so all the forces must add up to zero. So in this frame we have $$\left(\sum F_x\right) + m r \omega^2 = 0; \quad \sum F_y = 0.$$ It is not hard to see that these two sets of equations are entirely equivalent.