5.10 Circular motion

We have seen in earlier chapters that the acceleration of a body moving in a circle of radius $R$ with uniform speed $v$ is $v^{2} / R$ directed towards the centre. According to the second law, the force $f_{c}$ providing this acceleration is :
$f_{c}=\frac{m v^{2}}{R}$
where $m$ is the mass of the body. This force-directed forwards the centre is called the centripetal force. For a stone rotated in a circle by a string, the centripetal force is provided by the tension in the string. The centripetal force for the motion of a planet around the sun is the gravitational force on the planet due to the sun. For a car taking a circular turn on a horizontal road, the centripetal force is the force of friction.
The circular motion of a car on a flat and banked road gives an interesting application of the laws of motion.

Motion of a car on a level road

Three forces act on the car
(i) The weight of the car, mg
(ii) Normal reaction, $N$
(iii) Frictional force, $f$
As there is no acceleration in the vertical direction
$N-m g=0$
$N=m g$

The centripetal force required for circular motion is along the surface of the road, and is provided by the component of the contact force between road and the car tyres along the surface. This by definition is the frictional force. Note that it is the static friction that provides the centripetal acceleration. Static friction opposes the impending motion of the car moving away from the circle. Using equation $f_{s} \leq \mu_{s} N$ and $f_{c}=\frac{m v^{2}}{R}$ we get the result
$f=\frac{m v^{2}}{R} \leq \mu_{s} N$
$v^{2} \leq \frac{\mu_{s} R N}{m}=\mu_{\mathrm{s}} R g$
$[\because N=m g]$
which is independent of the mass of the car. This shows that for a given value of $\mu_{\mathrm{s}}$ and $R$, there is a maximum speed of circular motion of the car possible, namely
$v_{\max }=\sqrt{\mu_{\mathrm{s}} R g} \dots(5.18)$

Motion of a car on a banked road 

Consider a vehicle of mass $m$ is moving with speed $v$ on a banked road of radius $R$ as shown in the diagram. Let $\theta$ be the angle of banking. $\mathrm{N}$ is the normal reaction exerted on the vehicle by a banked road.
Let $\mathrm{f}$ be the frictional force between the road and the tyres of the vehicle.
From Free Body Diagram, we can say that,
Total Upward Force $=$ Total Downward Force
$N \cos \theta=m g+f \sin \theta$
$m g=N \cos \theta-f \sin \theta$……. (1)
Horizontal Component Nsin $\theta$ and f $\cos \theta$ will provide the necessary centripetal force.
$\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{N} \sin \theta+\mathrm{f} \cos \theta \ldots \ldots . .(2)$
On dividing (2) by (1), $\frac{\mathrm{v}^{2}}{\mathrm{Rg}}=\frac{\mathrm{N} \sin \theta+\mathrm{f} \cos \theta}{\mathrm{N} \cos \theta-\mathrm{f} \sin \theta}$
But frictional force, $\mathrm{f}=\mu \mathrm{N}$
where $\mu=$ coefficient of friction between road and tyres.
$\frac{\mathrm{v}^{2}}{\mathrm{Rg}}=\frac{\mathrm{N} \sin \theta+\mu \mathrm{N} \cos \theta}{\mathrm{N} \cos \theta-\mu \mathrm{N} \sin \theta}$
$\frac{\mathrm{v}^{2}}{\mathrm{Rg}}=\frac{\frac{\sin \theta}{\cos \theta}+\mu \frac{\cos \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}-\mu \frac{\sin \theta}{\cos \theta}}$
$\begin{aligned} &\therefore \frac{\mathrm{v}^{2}}{\mathrm{Rg}}=\frac{\mu+\tan \theta}{1-\mu \tan \theta} \\ &\mathrm{v}=\sqrt{\left(\frac{\mu+\tan \theta}{1-\mu \tan \theta}\right) \mathrm{Rg}} \end{aligned}$
This is the speed with which a vehicle can move safely on a rough circular banked roads.
From this equation, we can conclude that :
(1) For a rough plane surface, $\theta=0$
$\mathrm{v}=\sqrt{\mu \mathrm{Rg}}$
(2) For a smooth banked road, $\mu=0$
$\mathrm{v}=\sqrt{\mathrm{Rg} \tan \theta}$

Example 5.18: A circular track of radius $300 \mathrm{~m}$ is banked at an angle $\frac{\pi}{12}$ radian. If the coefficient of friction between wheel of a car and road is $0.2$, the maximum safe speed of car is:

Solution: Maximum velocity,
$\begin{aligned} &\mathrm{v}_{\max }=\left[\left(\frac{\mu+\tan \theta}{1-\mu \tan \theta}\right) \mathrm{rg}\right]{ }^{1 / 2} \\ &=\left[\left(\frac{0.2+\tan 15^{\circ}}{1-0.2 \tan 15^{\circ}}\right) 300 \times 9.8\right] \\ &\simeq 38 \mathrm{~ms}^{-1} \end{aligned}$

Example 5.19: A cyclist speeding at $18 \mathrm{~km} / \mathrm{h}$ on a level road takes a sharp circular turn of radius $3 \mathrm{~m}$ without reducing the speed. The coefficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?

Answer: On an unbanked road, frictional force alone can provide the centripetal force needed to keep the cyclist moving on a circular turn without slipping. If the speed is too large, or if the turn is too sharp (1.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by Eq. (5.18) :
$v^2 \leq \mu_s R g$
Now, $R=3 \mathrm{~m}, g=9.8 \mathrm{~m} \mathrm{~s}^{-2}, \mu_s=0.1$. That is, $\mu_s R g=2.94 \mathrm{~m}^2 \mathrm{~s}^{-2} . v=18 \mathrm{~km} / \mathrm{h}=5 \mathrm{~m} \mathrm{~s}^{-1}$;1.e., $v^2=25 \mathrm{~m}^2 \mathrm{~s}^{-2}$. The condition is not obeyed. The cyclist will slip while taking the circular turn.