In short

In a conical pendulum, the string tension resolves into two components: the vertical component balances the bob's weight (T\cos\theta = mg) and the horizontal component provides the centripetal force (T\sin\theta = mv^2/r). Dividing these gives \tan\theta = v^2/rg. The same equation governs the banking of roads: a road tilted at angle \theta provides centripetal force through the normal reaction, with the ideal speed given by v = \sqrt{rg\tan\theta}. When friction is included, the maximum and minimum safe speeds are v_{\max} = \sqrt{rg\,\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}} and v_{\min} = \sqrt{rg\,\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta}}.

Tie a small stone to a string, hold the other end in your hand, and whirl the stone in a horizontal circle above your head. The string does not hang straight down — it sweeps out a cone, with the stone tracing a steady circle at the base. You have just built a conical pendulum. The faster you spin the stone, the wider the cone opens. Slow down, and the cone narrows. There is a direct link between the speed of the stone and the angle the string makes with the vertical.

Now picture an autorickshaw taking a sharp curve on a flyover in Hyderabad. The flyover is not flat — it tilts inward, banked like a velodrome. That tilt does the same job as the string in your conical pendulum: it provides a sideways force that bends the autorickshaw's path into a circle. Without the tilt, the tyres would have to supply all the centripetal force through friction alone. On a wet monsoon evening, that friction might not be enough.

These two problems — the whirling stone and the banked road — share the same physics. In both cases, a force that is not horizontal gets resolved into a vertical component (balancing weight) and a horizontal component (providing centripetal acceleration). The mathematics is identical, and once you solve one, you have solved the other.

The conical pendulum

Setting up the problem

A bob of mass m is attached to one end of a string of length L. The other end is fixed to a point on the ceiling. The bob moves in a horizontal circle of radius r at constant speed v. The string makes a constant angle \theta with the vertical as it sweeps out a cone.

Assumptions: The string is massless and inextensible. Air resistance is negligible. The bob moves at constant speed in a horizontal circle (uniform circular motion).

Free body diagram of a conical pendulum Left: a conical pendulum showing the string of length L at angle theta from the vertical, bob tracing a horizontal circle of radius r at height h below the pivot. Right: the free body diagram of the bob showing tension T along the string, weight mg downward, and the resolved components T cos theta upward and T sin theta inward (horizontal). Setup pivot θ r h L m Free body diagram m mg T T cos θ T sin θ x (inward) y
Left: the conical pendulum — a bob of mass $m$ traces a horizontal circle of radius $r$ while the string of length $L$ sweeps a cone at angle $\theta$. Right: the free body diagram showing only two forces on the bob — tension $T$ along the string and weight $mg$ downward. The tension resolves into $T\cos\theta$ (vertical) and $T\sin\theta$ (horizontal, toward the centre).

The two forces on the bob

Only two forces act on the bob:

  1. Tension T in the string, directed along the string toward the pivot.
  2. Weight mg, directed straight downward.

There is no force pushing the bob outward — that "centrifugal force" feeling is what you experience in a rotating frame, not a real force in the lab frame. In the lab frame, the bob accelerates inward (centripetally), and the horizontal component of tension provides that acceleration.

Deriving the key results

The bob moves in a horizontal circle at constant speed. Its acceleration is centripetal — directed horizontally inward, toward the centre of the circle. There is no vertical acceleration (the bob stays at the same height).

Step 1. Resolve the tension into vertical and horizontal components.

The string makes angle \theta with the vertical. The component along the vertical is T\cos\theta. The component along the horizontal (pointing toward the centre of the circle) is T\sin\theta.

Why: when you decompose a force along directions that are parallel and perpendicular to the vertical, the adjacent component uses \cos and the opposite component uses \sin. The angle \theta is measured from the vertical, so the vertical component is T\cos\theta.

Step 2. Apply Newton's second law in the vertical direction.

The bob has no vertical acceleration, so the net vertical force is zero:

T\cos\theta = mg \tag{1}

Why: the upward pull of the string's vertical component must exactly balance the bob's weight. If it did not, the bob would accelerate up or down, and it would not stay in a horizontal circle.

Step 3. Apply Newton's second law in the horizontal direction.

The net horizontal force provides the centripetal acceleration v^2/r:

T\sin\theta = \frac{mv^2}{r} \tag{2}

Why: the only horizontal force is the inward component of tension. This must equal ma_c = mv^2/r for circular motion. No friction, no normal force — just the string pulling inward.

Step 4. Divide equation (2) by equation (1) to eliminate T.

\frac{T\sin\theta}{T\cos\theta} = \frac{mv^2/r}{mg}
\tan\theta = \frac{v^2}{rg} \tag{3}

Why: dividing removes both T and m. The angle depends only on the speed, the radius, and g — not on the mass of the bob. A heavy bob and a light bob at the same speed and radius make the same angle. This is analogous to how the period of a simple pendulum is independent of mass.

Step 5. Express the radius in terms of the string length.

From the geometry of the cone: r = L\sin\theta and the height h = L\cos\theta.

Substituting r = L\sin\theta into equation (3):

\tan\theta = \frac{v^2}{L\sin\theta \cdot g}
\frac{\sin\theta}{\cos\theta} = \frac{v^2}{Lg\sin\theta}
\sin^2\theta = \frac{v^2 \cos\theta}{Lg}

Why: multiplying both sides by \sin\theta and using \tan\theta = \sin\theta/\cos\theta gives a relation between \theta, v, L, and g. This lets you find the angle for a given speed, or vice versa.

Step 6. Find the time period of revolution.

The bob travels a circle of circumference 2\pi r at speed v. The time period is:

T_{\text{period}} = \frac{2\pi r}{v}

From equation (3): v^2 = rg\tan\theta, so v = \sqrt{rg\tan\theta}.

T_{\text{period}} = \frac{2\pi r}{\sqrt{rg\tan\theta}} = 2\pi\sqrt{\frac{r}{g\tan\theta}}

Using r = L\sin\theta and \tan\theta = \sin\theta/\cos\theta:

T_{\text{period}} = 2\pi\sqrt{\frac{L\sin\theta \cdot \cos\theta}{g\sin\theta}} = 2\pi\sqrt{\frac{L\cos\theta}{g}}
\boxed{T_{\text{period}} = 2\pi\sqrt{\frac{L\cos\theta}{g}}} \tag{4}

Why: this is remarkably similar to the time period of a simple pendulum (2\pi\sqrt{L/g}). The difference is the \cos\theta factor. As \theta increases (cone opens wider), \cos\theta decreases, and the period gets shorter — the bob revolves faster. At \theta \to 0 (almost vertical string), the conical pendulum period approaches the simple pendulum period.

You can also write this as T_{\text{period}} = 2\pi\sqrt{h/g}, where h = L\cos\theta is the height of the bob below the pivot. The period depends only on the vertical height, not directly on the string length or the cone angle separately.

Banking of roads

Why flat roads are dangerous on curves

When a car takes a curve on a flat road, the only force bending its path into a circle is static friction between the tyres and the road surface. On the Mumbai-Pune Expressway, the curves are designed for speeds of 80-100 km/h. If the road were perfectly flat, the frictional force needed would be:

f = \frac{mv^2}{r}

On a rainy day, \mu_s drops to perhaps 0.3 or lower. If \frac{v^2}{rg} > \mu_s, the car skids outward. This is why flat curves are speed-limited — and why engineers bank the road.

The banked road without friction

Banking means tilting the road surface inward, so that the outer edge is higher than the inner edge. On a banked road, the normal force N from the road surface is no longer vertical — it tilts inward, and its horizontal component can provide the centripetal force.

Free body diagram of a car on a banked road (no friction) Left: a cross-section of a banked road showing a car on an inclined surface tilted at angle theta. Right: the free body diagram of the car with normal force N perpendicular to the surface, weight mg downward, and N resolved into N cos theta (vertical) and N sin theta (horizontal, toward centre). Cross-section of banked road θ car inner edge outer edge to centre Free body diagram mg N N cos θ N sin θ toward centre up
Left: cross-section of a banked curve. The road tilts inward at angle $\theta$, making the outer edge higher. Right: FBD of the car. The normal force $N$ is perpendicular to the road surface, not vertical. Its horizontal component $N\sin\theta$ points toward the centre of the curve, providing centripetal force. Its vertical component $N\cos\theta$ balances the weight.

Deriving the optimal banking angle

Consider a car of mass m moving at speed v along a banked curve of radius r. The road is banked at angle \theta. For the frictionless case, only two forces act on the car: its weight mg downward and the normal force N perpendicular to the road surface.

Step 1. Apply Newton's second law in the vertical direction.

The car does not accelerate vertically:

N\cos\theta = mg \tag{5}

Why: the vertical component of the normal force must balance the car's weight. This is identical in structure to the conical pendulum's T\cos\theta = mg — the normal force here plays the same role as the tension there.

Step 2. Apply Newton's second law in the horizontal direction.

The horizontal component of N provides the centripetal force:

N\sin\theta = \frac{mv^2}{r} \tag{6}

Why: the only horizontal force is the inward component of the normal reaction. Friction is absent in this ideal case, so N\sin\theta alone bends the car's path into a circle.

Step 3. Divide equation (6) by equation (5).

\frac{N\sin\theta}{N\cos\theta} = \frac{mv^2/r}{mg}
\boxed{\tan\theta = \frac{v^2}{rg}} \tag{7}

Why: the mass m and the normal force N both cancel. The banking angle depends only on the intended speed, the radius of the curve, and g. This is the same equation as the conical pendulum — equation (3). The physics is identical: resolve a tilted contact force into centripetal and weight-balancing components.

This is the optimal banking angle — the angle at which a car can take the curve at speed v with zero friction. At exactly this speed, the road pushes the car inward with exactly the right force, and the tyres do not need to grip sideways at all. Even on a perfectly smooth ice-covered road, the car would navigate the curve safely at this speed.

The ideal speed for a given banking angle and radius is:

v_{\text{ideal}} = \sqrt{rg\tan\theta}

On Indian national highways, curves on elevated sections (flyovers, expressway ramps) are typically banked at 5° to 15°. The Mumbai-Pune Expressway has curves banked for speeds around 80 km/h.

Banking with friction — the real-world case

The frictionless result gives one specific speed for each banking angle. But real cars drive at a range of speeds. Friction allows the car to go somewhat faster or slower than the ideal speed without skidding.

Setting up the forces

When the car goes faster than the ideal speed, it tends to slide outward (up the slope). Friction acts down the slope (inward), adding to the centripetal force.

When the car goes slower than the ideal speed, it tends to slide inward (down the slope). Friction acts up the slope (outward), reducing the net centripetal force to match the smaller v^2/r needed.

FBD of a car on a banked road with friction for the fast-car case Free body diagram showing the car on a banked road going faster than ideal speed. Normal force N perpendicular to road, weight mg downward, friction f pointing down the slope (inward). The components are resolved along vertical and horizontal axes. Faster than ideal speed — friction acts down the slope road surface mg N f to centre N cos θ + f sin θ N sin θ + f cos θ
When the car goes faster than the ideal speed, it tends to slide up the slope. Friction $f$ acts down the slope (toward the centre). Both friction and the normal force contribute horizontal components toward the centre, giving a larger centripetal force and allowing higher speeds.

Deriving maximum speed (friction assists banking)

When the car moves at the maximum safe speed, friction is at its limit: f = \mu_s N, acting down the slope (toward the centre of the curve).

Step 1. Resolve all forces along the vertical direction (no vertical acceleration):

N\cos\theta - \mu_s N\sin\theta = mg
N(\cos\theta - \mu_s\sin\theta) = mg \tag{8}

Why: N\cos\theta acts upward. Friction acts down the slope, so its vertical component f\sin\theta = \mu_s N\sin\theta acts downward. The vertical equilibrium requires their difference to equal mg.

Step 2. Resolve all forces along the horizontal direction (centripetal):

N\sin\theta + \mu_s N\cos\theta = \frac{mv_{\max}^2}{r}
N(\sin\theta + \mu_s\cos\theta) = \frac{mv_{\max}^2}{r} \tag{9}

Why: both N\sin\theta (from banking) and f\cos\theta = \mu_s N\cos\theta (from friction) point horizontally toward the centre. They add up to provide the centripetal force.

Step 3. Divide equation (9) by equation (8) to eliminate N and m:

\frac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta} = \frac{v_{\max}^2}{rg}

Divide numerator and denominator of the left side by \cos\theta:

\boxed{v_{\max} = \sqrt{rg\,\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}}} \tag{10}

Why: when \mu_s = 0, this reduces to v = \sqrt{rg\tan\theta} — the frictionless ideal speed. When \theta = 0 (flat road), it gives v_{\max} = \sqrt{\mu_s rg}, which is the familiar result for the maximum speed on a flat curve with friction alone.

Deriving minimum speed (friction opposes sliding inward)

When the car moves too slowly, gravity's component down the slope exceeds the centripetal force needed. The car tends to slide inward (down the slope), so friction acts up the slope (outward).

The same analysis with friction reversed gives:

N(\cos\theta + \mu_s\sin\theta) = mg \tag{11}
N(\sin\theta - \mu_s\cos\theta) = \frac{mv_{\min}^2}{r} \tag{12}

Why: now friction's vertical component helps support the weight (acts upward), and friction's horizontal component opposes the centripetal direction (acts outward). Both sign changes come from reversing the friction direction.

Dividing equation (12) by equation (11):

\boxed{v_{\min} = \sqrt{rg\,\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta}}} \tag{13}

Why: this gives the slowest speed at which the car can navigate the curve without sliding inward. If \tan\theta \leq \mu_s, the expression under the square root is zero or negative — meaning friction is strong enough to hold the car stationary on the banked surface, so there is no minimum speed constraint.

The safe speed range for a banked curve with friction is v_{\min} \leq v \leq v_{\max}.

Worked examples

Example 1: A conical pendulum — finding speed and period

A stone of mass 200 g is attached to a string of length 1 m. The stone moves in a horizontal circle with the string making an angle of 30° with the vertical. Find (a) the speed of the stone and (b) the time period of revolution. Take g = 9.8 m/s².

Conical pendulum with L = 1 m and theta = 30 degrees A conical pendulum with string length 1 metre at 30 degrees from vertical. The bob traces a horizontal circle of radius 0.5 m at a height of 0.866 m below the pivot. Key dimensions and forces are labelled. pivot 30° r = 0.5 m h = 0.866 m L = 1 m T mg v = ?
The conical pendulum with $L = 1$ m and $\theta = 30°$. The radius of the horizontal circle is $r = L\sin 30° = 0.5$ m, and the height is $h = L\cos 30° = 0.866$ m.

Step 1. Find the radius of the circular path.

r = L\sin\theta = 1 \times \sin 30° = 0.5 \text{ m}

Why: from the geometry of the cone, the radius is the horizontal distance from the axis to the bob.

Step 2. Find the speed using \tan\theta = v^2/(rg).

v^2 = rg\tan\theta = 0.5 \times 9.8 \times \tan 30°
v^2 = 0.5 \times 9.8 \times 0.5774 = 2.829 \text{ m}^2/\text{s}^2
v = \sqrt{2.829} = 1.68 \text{ m/s}

Why: this is the speed at which the bob must move for the string to maintain exactly 30° with the vertical. At a lower speed, the angle would be smaller; at a higher speed, larger.

Step 3. Find the time period using equation (4).

T_{\text{period}} = 2\pi\sqrt{\frac{L\cos\theta}{g}} = 2\pi\sqrt{\frac{1 \times \cos 30°}{9.8}}
T_{\text{period}} = 2\pi\sqrt{\frac{0.866}{9.8}} = 2\pi\sqrt{0.0884}
T_{\text{period}} = 2\pi \times 0.2973 = 1.87 \text{ s}

Why: you can verify this independently. The circumference is 2\pi r = 2\pi \times 0.5 = 3.14 m. At speed 1.68 m/s, the time to go around once is 3.14/1.68 = 1.87 s. The two calculations agree.

Result: The speed of the stone is 1.68 m/s, and the period of revolution is 1.87 s.

What this shows: Even a modest angle of 30° requires a speed of only 1.68 m/s — about the speed of a brisk walk. The conical pendulum is a gentle system. Notice that the mass (200 g) never appeared in the calculation — the speed and period depend only on L, \theta, and g.

Example 2: A banked road on the Mumbai-Pune Expressway

A curve on a highway is banked at 15° with a radius of curvature of 100 m. (a) Find the ideal speed (no friction needed). (b) If the coefficient of static friction between tyres and road is \mu_s = 0.2, find the maximum safe speed and the minimum safe speed. Take g = 9.8 m/s².

FBD for a car on a road banked at 15 degrees with friction Free body diagram showing a car on a banked road at 15 degrees. For the maximum speed case: weight mg acts downward, normal N acts perpendicular to road surface, and friction f acts down the slope (toward centre). The angle theta is 15 degrees, radius of curve is 100 m. 15° mg N f = μₛN θ = 15° r = 100 m μₛ = 0.2 g = 9.8 m/s² Given to centre
FBD for a car on a 15° banked curve (maximum speed case). The normal force $N$ is perpendicular to the road. Friction $f = \mu_s N$ acts down the slope, assisting the centripetal force. The data box lists all given values.

Part (a): Ideal speed (no friction).

v_{\text{ideal}} = \sqrt{rg\tan\theta} = \sqrt{100 \times 9.8 \times \tan 15°}
v_{\text{ideal}} = \sqrt{100 \times 9.8 \times 0.2679} = \sqrt{262.5}
v_{\text{ideal}} = 16.2 \text{ m/s} \approx 58.3 \text{ km/h}

Why: at this speed, the normal force alone provides exactly the right centripetal force. No friction is needed at all — the car could drive on ice at this speed and still make the turn safely.

Part (b): Maximum safe speed.

Apply equation (10):

v_{\max} = \sqrt{rg\,\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}}
v_{\max} = \sqrt{100 \times 9.8 \times \frac{0.2679 + 0.2}{1 - 0.2 \times 0.2679}}
v_{\max} = \sqrt{980 \times \frac{0.4679}{0.9464}}
v_{\max} = \sqrt{980 \times 0.4943} = \sqrt{484.4}
v_{\max} = 22.0 \text{ m/s} \approx 79.2 \text{ km/h}

Why: friction adds to the centripetal force, so the car can go about 36% faster than the ideal speed before it starts to skid outward. The factor \frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta} always exceeds \tan\theta when \mu_s > 0.

Part (c): Minimum safe speed.

Apply equation (13):

v_{\min} = \sqrt{rg\,\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta}}
v_{\min} = \sqrt{100 \times 9.8 \times \frac{0.2679 - 0.2}{1 + 0.2 \times 0.2679}}
v_{\min} = \sqrt{980 \times \frac{0.0679}{1.0536}}
v_{\min} = \sqrt{980 \times 0.0645} = \sqrt{63.2}
v_{\min} = 7.95 \text{ m/s} \approx 28.6 \text{ km/h}

Why: since \tan 15° = 0.268 > \mu_s = 0.2, the car cannot remain stationary on this slope — friction alone is not strong enough to prevent it from sliding inward if it is going too slowly. It must maintain at least 28.6 km/h to stay on the road.

Result: The ideal speed is 58.3 km/h. With friction (\mu_s = 0.2), the safe speed range is 28.6 km/h to 79.2 km/h.

What this shows: Banking dramatically extends the safe speed range. A flat road with \mu_s = 0.2 and r = 100 m would allow a maximum speed of only \sqrt{\mu_s rg} = \sqrt{0.2 \times 100 \times 9.8} = 14.0 m/s = 50.4 km/h. The 15° banking pushes the maximum speed up to 79.2 km/h — a 57% increase. This is why highway engineers bank every significant curve.

Common confusions

If the derivations above made sense and you can solve the two examples, you understand the core physics. What follows is for readers preparing for JEE Advanced — the connection between the conical pendulum and SHM, and a careful look at the limiting cases of banking with friction.

The conical pendulum and the simple pendulum

The conical pendulum's time period is T = 2\pi\sqrt{L\cos\theta/g}. The simple pendulum's time period (small angle) is T_0 = 2\pi\sqrt{L/g}. The ratio is:

\frac{T}{T_0} = \sqrt{\cos\theta}

As \theta \to 0, the cone collapses to a straight line, and the conical pendulum approaches a simple pendulum swinging in a tiny arc. This is not a coincidence — both systems are governed by the same restoring mechanism (the component of tension perpendicular to the equilibrium direction), and the conical pendulum at small angles is essentially SHM projected onto a plane.

For \theta = 30°: T/T_0 = \sqrt{\cos 30°} = \sqrt{0.866} = 0.931. The conical pendulum is about 7% faster than the simple pendulum with the same string length.

Tension in the string

From T\cos\theta = mg:

T = \frac{mg}{\cos\theta}

The tension is always greater than the weight (since \cos\theta < 1 for \theta > 0). As \theta increases, the tension grows. At \theta = 60°, T = mg/\cos 60° = 2mg — the string pulls with twice the bob's weight. At \theta \to 90° (a horizontal string), T \to \infty, which is physically impossible. You cannot make a conical pendulum with a perfectly horizontal string — you would need infinite speed and infinite tension. This sets a physical upper limit on how wide the cone can open.

When does the minimum speed vanish?

From equation (13), v_{\min} = 0 when \tan\theta \leq \mu_s. This means:

\theta \leq \arctan(\mu_s)

For \mu_s = 0.2: \theta \leq \arctan(0.2) = 11.3°. If the banking angle is less than 11.3°, friction can hold a stationary car on the slope, and there is no minimum speed — you can stop on the curve without sliding inward.

For \mu_s = 0.5 (dry tyres on dry asphalt): \theta \leq \arctan(0.5) = 26.6°. Most real roads are banked well below this, so on dry roads, the minimum speed constraint rarely applies. It becomes relevant on wet or icy surfaces where \mu_s drops sharply.

What if \mu_s \tan\theta = 1?

In equation (10), the denominator 1 - \mu_s\tan\theta appears. If \mu_s\tan\theta = 1, the denominator is zero and v_{\max} \to \infty. This happens when \theta = \arctan(1/\mu_s).

For \mu_s = 0.5: \theta = \arctan(2) = 63.4°. At this extreme banking angle, friction combined with the normal force can theoretically sustain any speed — the car will never skid outward. Of course, no real road is banked at 63°; this is a mathematical limit, not an engineering prescription. But it explains why velodrome bicycle tracks, banked at 40-45°, allow cyclists to ride at extremely high speeds without sliding outward.

Connecting the two problems

The conical pendulum and the banked road are the same free body diagram with different labels:

Conical pendulum Banked road (no friction)
Tension T Normal force N
Bob mass m Car mass m
String angle \theta from vertical Banking angle \theta from horizontal... wait — from vertical
Radius r = L\sin\theta Radius r (given)
\tan\theta = v^2/(rg) \tan\theta = v^2/(rg)

The equations are identical because the physics is identical: a single tilted force being resolved into a centripetal component and a weight-balancing component. Once you recognise this pattern, you can apply it to any problem where a contact force is angled — a satellite in a banked orbit, a cyclist leaning into a turn, a fighter jet in a banked turn.

Where this leads next