Physics · Ch 4 — Laws of Motion

Circular Motion

4.10

Circular Motion

The Centripetal Force Requirement

From Chapter 3, we already know that a body moving in a circle of radius $R$ with uniform speed $v$ experiences an acceleration $v^2/R$ directed toward the centre. Newton's second law then tells us that a force must be providing this acceleration. That force is

f_c = \frac{mv^2}{R}

where $m$ is the mass of the body. This inward-directed force is called the centripetal force.

The physical source of the centripetal force changes with the situation. For a stone whirled in a circle by a string, the tension in the string provides it. For a planet orbiting the Sun, the gravitational pull of the Sun on the planet does the job. For a car taking a circular turn on a horizontal road, the centripetal force comes from friction between the tyres and the road.

Two particularly instructive applications are the motion of a car on a level road and on a banked road. These show how the laws of motion combine with geometry to determine safe speeds.

Motion of a Car on a Level Road

Three forces act on the car as it moves on a horizontal circular track:

The weight of the car, $mg$ , acting vertically downward.
The normal reaction from the road, $N$ , acting vertically upward.
The frictional force, $f$ , acting horizontally along the surface of the road.

Since there is no acceleration in the vertical direction, the vertical forces must balance:

N - mg = 0 \quad \Rightarrow \quad N = mg

The centripetal force needed for circular motion acts along the surface of the road, toward the centre of the circle. This force is provided by the component of the contact force between the road and the tyres that lies along the surface — that is, by the frictional force. Static friction is what opposes the impending motion of the car away from the circle.

Watch out

It is static friction, not kinetic friction, that provides the centripetal force here. The car's tyres are not sliding relative to the road; they are rolling. The friction that prevents slipping at the contact point is static friction.

The maximum value of static friction is $f_s^{\text{max}} = \mu_s N$ , where $\mu_s$ is the coefficient of static friction. For the car to stay on its circular path without slipping, the required centripetal force must not exceed this maximum:

\frac{mv^2}{R} \leq \mu_s N

Using $N = mg$ , this becomes

\frac{mv^2}{R} \leq \mu_s mg

Cancelling $m$ from both sides gives

\frac{v^2}{R} \leq \mu_s g \quad \Rightarrow \quad v^2 \leq \mu_s R g

The maximum safe speed on a level road is therefore

v_{\text{max}} = \sqrt{\mu_s R g}

Notice that this result is independent of the mass of the car. For a given radius of turn $R$ and coefficient of friction $\mu_s$ , every vehicle has the same maximum safe speed.

v_{\text{max}} = \sqrt{\mu_s R g}

Motion of a Car on a Banked Road

If the road is banked — that is, tilted at an angle $\theta$ to the horizontal — the normal reaction itself has a horizontal component that can contribute to the centripetal force. This reduces the reliance on friction and allows a higher safe speed.

Consider a car moving on a road banked at angle $\theta$ . The forces acting are the same three — weight $mg$ , normal reaction $N$ , and friction $f$ — but now $N$ is not vertical and $f$ is not horizontal. The road surface is tilted, so both $N$ and $f$ have vertical and horizontal components.

Force Balance in the Vertical Direction

Since there is no acceleration vertically, the net vertical force must be zero. The vertical component of $N$ is $N\cos\theta$ (upward), the vertical component of $f$ is $f\sin\theta$ (also upward if friction acts up the slope), and the weight $mg$ acts downward. Hence

N\cos\theta = mg + f\sin\theta

Force Balance in the Horizontal Direction

The centripetal force required is $mv^2/R$ , directed horizontally toward the centre of the circle. The horizontal components of $N$ and $f$ together provide this:

N\sin\theta + f\cos\theta = \frac{mv^2}{R}

Maximum Permissible Speed

Friction cannot exceed its maximum value: $f \leq \mu_s N$ . To find the maximum possible speed $v_{\text{max}}$ , we take the limiting case $f = \mu_s N$ . Substituting this into the two balance equations gives

N\cos\theta = mg + \mu_s N\sin\theta \qquad(4.20a)

N\sin\theta + \mu_s N\cos\theta = \frac{mv^2}{R} \qquad(4.20b)

From equation (4.20a), we can solve for $N$ :

N(\cos\theta - \mu_s\sin\theta) = mg \quad \Rightarrow \quad N = \frac{mg}{\cos\theta - \mu_s\sin\theta}

Substitute this expression for $N$ into equation (4.20b):

\frac{mg}{\cos\theta - \mu_s\sin\theta}(\sin\theta + \mu_s\cos\theta) = \frac{mv^2}{R}

The mass $m$ cancels. Rearranging,

v^2 = Rg\,\frac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta}

Dividing numerator and denominator by $\cos\theta$ gives a cleaner form:

v^2 = Rg\,\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}

Therefore the maximum permissible speed on a banked road is

v_{\text{max}} = \sqrt{Rg\left(\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}\right)}

v_{\text{max}} = \sqrt{Rg\left(\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}\right)}

Comparing this with $v_{\text{max}} = \sqrt{\mu_s R g}$ for a level road, we see that banking allows a higher speed for the same radius and coefficient of friction.

Optimum Speed (No Friction Needed) …

Figure 4.14Circular motion of a car on (a) a level road, (b) a banked road.

Fig. 4.14 — Circular motion of a car on (a) a level road, (b) a banked road.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

The figure places two scenarios side by side. In panel (a), a car moves on a flat, level road and follows a circular path of radius $R$ centred at $O$ . The car’s velocity is tangential to the circle, and its acceleration $a = v^2/R$ points radially inward toward $O$ . Two vertical forces act: the normal reaction $N$ upward and the weight $mg$ downward. These cancel vertically, so they do not affect the horizontal motion. The only horizontal force is static friction $f$ , which acts toward the centre of the circle. This friction provides the necessary centripetal force.

Panel (b) shows the same car on a road that is banked at an angle $\theta$ to the horizontal. The normal reaction $N$ is no longer vertical — it is perpendicular to the road surface. The weight $mg$ still acts straight down. Friction $f$ acts along the road surface, parallel to it. Both $N$ and $f$ are resolved into horizontal and vertical components: $N\cos\theta$ and $N\sin\theta$ for the normal, and $f\cos\theta$ and $f\sin\theta$ for friction. The horizontal components together supply the centripetal force, while the vertical components balance the weight.

The key physical idea is that banking reduces or even eliminates the need for friction to provide the centripetal force. On a level road, friction is the sole provider; on a banked road, a component of the normal reaction itself points inward, so the car can negotiate the curve at a certain “design speed” even if the road is icy (friction zero).

N\sin\theta + f\cos\theta = \frac{mv^2}{R}

Here $m$ is the mass of the car, $v$ its speed, $R$ the radius of the circular path, $\theta$ the banking angle, $N$ the normal reaction, and $f$ the static friction (which can be at most $\mu_s N$ , where $\mu_s$ is the coefficient of static friction). The first equation is the horizontal (centripetal) force balance; the second is the vertical force balance. …

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