12.1 Motion along curved path  (Page 4/4)

Skidding down the bank

In this case, the velocity of the vehicle is less than threshold speed " $\sqrt{rg\tan \theta }$ ". Friction acts "up" across the bank. There are three forces acting on the vehicle (i) its weight "mg" (ii) normal force (N) due to the bank surface and (iii) static friction " $f_s$ ", acting up the bank. The free body diagram is as shown here.

$$\sum F_x\Rightarrow N\sin \theta -f_S\cos \theta =\frac{m{v}^2}{r}$$

$$\sum F_y\Rightarrow N\cos \theta +f_S\sin \theta =mg$$

Skidding up the bank

In this case, the velocity of the vehicle is greater than threshold speed. Friction acts "down" across the bank. There are three forces acting on the vehicle (i) its weight "mg" (ii) normal force (N) due to the bank surface and (iii) static friction " $f_s$ ", acting down the bank. The free body diagram is as shown here.

$$\sum F_x\Rightarrow N\sin \theta +f_S\cos \theta =\frac{m{v}^2}{r}$$

$$\sum F_y\Rightarrow N\cos \theta -f_S\sin \theta =mg$$

Maximum speed along the banked road

In previous section, we discussed various aspects of banking. In this section, we seek to find the maximum speed with which a banked curve can be negotiated. We have seen that banking, while preventing upward skidding, creates situation in which the vehicle can skid downward at lower speed.

The design of bank, therefore, needs to consider both these aspects. Actually, roads are banked with a small angle of inclination only. It is important as greater angle will induce tendency for the vehicle to overturn. For small inclination of the bank, the tendency of the vehicle to slide down is ruled out as friction between tyres and road is usually much greater to prevent downward skidding across the road.

In practice, it is the skidding "up" across the road that is the prime concern as threshold speed limit can be breached easily. The banking supplements the provision of centripetal force, which is otherwise provided by the friction on a flat road. As such, banking can be seen as a mechanism either (i) to increase the threshold speed limit or (ii) as a safety mechanism to cover the risk involved due to any eventuality like flattening of tyres or wet roads etc. In fact, it is the latter concern that prevails.

In the following paragraph, we set out to determine the maximum speed with which a banked road can be negotiated. It is obvious that maximum speed corresponds to limiting friction that acts in the downward direction as shown in the figure.

Force analysis in the vertical direction :

$$\begin{array}{l}N\cos \theta -{\mu }_{s}N\sin \theta =mg\\ \Rightarrow N(\cos \theta -{\mu }_s\sin \theta )=mg\end{array}$$

Force analysis in the horizontal direction :

$$\begin{array}{l}N\sin \theta +{\mu }_{s}N\cos \theta =\frac{m{v}^2}{r}\\ \Rightarrow N(\sin \theta +{\mu }_s\cos \theta )=\frac{m{v}^2}{r}\end{array}$$

Taking ratio of two equations, we have :

$$\begin{array}{l}\Rightarrow \frac{g(\sin \theta +{\mu }_s\cos \theta )}{(\cos \theta -{\mu }_s\sin \theta )}=\frac{v^2}{r}\\ \Rightarrow v^2=rg\frac{(\sin \theta +{\mu }_s\cos \theta )}{(\cos \theta -{\mu }_s\sin \theta )}\\ \Rightarrow v=\surd \left\{rg\frac{(\tan \theta +{\mu }_s)}{(1-{\mu }_s\tan \theta )}\right\}\end{array}$$

Bending by a cyclist

We have seen that a cyclist bends towards the center in order to move along a circular path. Like in the case of car, he could have depended on the friction between tires and the road. But then he would be limited by the speed. Further, friction may not be sufficient as contact surface is small. We can also see "bending" of cyclist at greater speed as an alternative to banking used for four wheeled vehicles, which can not be bent.

The cyclist increases speed without skidding by leaning towards the center of circular path. The sole objective of bending here is to change the direction and magnitude of normal force such that horizontal component of the normal force provides for the centripetal force, whereas vertical component balances the "cycle and cyclist" body system.

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

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

Taking ratio,

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