# 12.1 Motion along curved path  (Page 2/4)

 Page 2 / 4

Important to note here is that we are not considering friction in the forward or actual direction of motion, but perpendicular to actual direction (side way). There is no motion in the side way direction, if there is no side way skidding of the car. In that case, the friction is static friction ( ${f}_{s}$ ) and has not exceeded maximum or limiting friction ( ${F}_{s}$ ). Thus,

$\begin{array}{l}{f}_{s}\le {F}_{s}\\ ⇒{f}_{s}\le {\mu }_{s}N\end{array}$

There is no motion in vertical direction. Hence,

$\begin{array}{l}N=mg\end{array}$

Combining two equations, we have :

$\begin{array}{l}⇒{f}_{s}\le {\mu }_{s}mg\end{array}$

where “m” is the combined mass of the car and the passengers.

In the limiting situation, when the car is about to skid away, the friction force is equal to the maximum static friction, meeting the requirement of centripetal force required for circular motion,

$\begin{array}{l}⇒\frac{m{v}^{2}}{r}={\mu }_{s}mg\\ ⇒v=\surd \left({\mu }_{s}rg\right)\end{array}$

We note following points about the condition as put on the speed of the car :

• There is a limiting or maximum speed of car to ensure that the car moves along curved path without skidding.
• If the limiting condition with regard to velocity is not met ( $v\le \surd \left({\mu }_{s}rg\right)$ ), then the car will skid away.
• The limiting condition is independent of the mass of the car.
• If friction between tires and the road is more, then we can negotiate a curved path with higher speed and vice-versa. This explains why we drive slow on slippery road.
• Smaller the radius of curvature, smaller is the limiting speed. This explains why sharper turn (smaller radius of curvature) is negotiated with smaller speed.

There are three additional aspects of negotiating a curve. First, what if, we want to negotiate curve at a higher speed. Second, how to make driving safer without attracting limiting conditions as tires may have been flattened (whose grooves have flattened), or friction may decrease due to any other reasons like rain or mud. Third, we want to avoid sideway friction to prolong life of the tires. The answer to these lie in banking of the curved road.

For banking, one side of the road is elevated from horizontal like an incline or wedge. In this case, the component of normal force in horizontal direction provides the centripetal force as required for the motion along the curved path. On the other hand, component of normal force in vertical direction balances the weight of the vehicle. It is clear that magnitude of normal reaction between road and vehicle is greater than the weight of the vehicle.

Here, we compute the relation between the angle of banking (which is equal to the angle of incline) for a given speed and radius of curvature as :

$\begin{array}{l}N\mathrm{cos}\theta =mg\end{array}$

and

$\begin{array}{l}N\mathrm{sin}\theta =\frac{m{v}^{2}}{r}\end{array}$

Taking ratio,

$\begin{array}{l}⇒\mathrm{tan}\theta =\frac{{v}^{2}}{rg}\\ ⇒v=\surd \left(rg\mathrm{tan}\theta \right)\end{array}$

This expression represents the speed at which the vehicle does not skid (up or down) along the banked road for the given angle of inclination (θ). It means that centripetal force is equal to the resultant of the system of forces acting on the vehicle. Importantly, there is no friction involved in this consideration for the circular motion of the vehicle.

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