Horizontal Circular Motion

A bob on a string moving in a horizontal circle

Horizontal circular motion — key situations: conical pendulum, banked curve (no friction), and banked curve with friction.

Tie a ball to a string and spin it above your head: the string traces a cone and the ball sweeps a horizontal circle at constant height. The string pulls the ball upward (supporting it against gravity) and inward (providing centripetal force). These two jobs are done by a single tension force at an angle — and resolving that tension into vertical and horizontal components lets you calculate the speed, radius, and period of the orbit.

Two forces act on the bob: tension $T$ along the string, and weight $mg$ downward. Resolving the tension into its two components:

• Vertical: $T\cos\theta = mg$ (balances weight, so the bob moves neither up nor down)
• Horizontal: $T\sin\theta = \dfrac{mv^2}{r}$ (provides the centripetal force)

Dividing the horizontal equation by the vertical equation eliminates $T$:

This is the fundamental relationship for a conical pendulum. Solving for speed: $v = \sqrt{rg\tan\theta}$.

The period (time for one complete revolution) uses $v = 2\pi r/T_\text{period}$ and $r = L\sin\theta$:

Conical pendulum: $r = L\sin\theta$; vertical $T\cos\theta = mg$ (tension $= mg/\cos\theta$); horizontal $T\sin\theta = mv^2/r$ → $\tan\theta = v^2/(rg)$; period $T = 2\pi\sqrt{L\cos\theta/g}$. Increasing speed widens the angle $\theta$.

Pause — copy the highlighted conical pendulum equations into your book before moving on.

When the normal force alone provides the centripetal force component

We just saw that the conical pendulum uses $\tan\theta = v^2/(rg)$ from resolving tension. That raises a question: do roads use the same idea — can banking replace the need for friction? This card answers it → yes; at design speed, $N\sin\theta$ provides all centripetal force and $v_\text{design} = \sqrt{rg\tan\theta}$.

A banked curve is a road or track inclined at angle $\theta$ to the horizontal. When a vehicle travels at exactly the design speed, the normal force $N$ from the road surface provides precisely the required centripetal force — no friction is needed.

Resolving the normal force:

• Vertical: $N\cos\theta = mg$ (balances weight)
• Horizontal: $N\sin\theta = \dfrac{mv^2}{r}$ (provides centripetal force)

Dividing horizontal by vertical eliminates $N$:

This is mathematically identical to the conical pendulum equation. Solving for the design speed:

Banked curve (no friction): $N\cos\theta = mg$; $N\sin\theta = mv^2/r$; dividing → $\tan\theta = v^2/(rg)$. Design speed $v_\text{design} = \sqrt{rg\tan\theta}$. Normal force $N = mg/\cos\theta > mg$.

Add the highlighted banking equations and the $N > mg$ result to your notes before the check below.

Friction acts up or down the slope depending on speed

We just saw that at the design speed, no friction is needed. That raises a question: what happens when the car goes faster or slower than design speed? This card answers it → friction acts down the slope (too fast) or up the slope (too slow), giving $v_\max = \sqrt{rg(\tan\theta + \mu_s)/(1 - \mu_s\tan\theta)}$.

When a vehicle travels at a speed different from the design speed, friction is required to prevent sliding. The direction of friction depends on the relationship between actual speed and design speed:

• $v > v_\text{design}$ (too fast): The vehicle tends to slide up the slope (toward the outer edge). Friction acts down the slope, adding to the centripetal force.
• $v < v_\text{design}$ (too slow): The vehicle tends to slide down the slope (toward the inner edge). Friction acts up the slope, but this reduces the net centripetal force.

Maximum Speed (Friction Acting Down the Slope)

At maximum speed, friction is at its limiting value $F_f = \mu_s N$ acting down the slope. Resolving forces:

Perpendicular to surface: $N = mg\cos\theta + \dfrac{mv^2_\max}{r}\sin\theta$

Along the horizontal (centripetal direction):

After algebraic manipulation:

Too fast ($v > v_\text{design}$): friction acts DOWN the slope. Too slow ($v < v_\text{design}$): friction acts UP. Maximum speed: $v_\max = \sqrt{rg(\tan\theta + \mu_s)/(1 - \mu_s\tan\theta)}$ — derive from force diagrams; do not just memorise.

Pause — write the highlighted friction-direction rules and the $v_\max$ formula into your book before moving on.

The limiting case: no banking, friction only

We just saw how banking raises the maximum safe speed through the normal force. That raises a question: what is the situation on an ordinary flat road with no banking at all? This card answers it → only friction available; $v_\max = \sqrt{\mu_s g r}$, which is independent of mass.

On a flat (unbanked) horizontal curve, the only horizontal force available to provide centripetal force is static friction between the tyres and the road. At the maximum speed before slipping:

Since $F_c = mv^2_\max/r$, substituting:

Flat curve: $F_c = \mu_s mg$, so $v_\max = \sqrt{\mu_s g r}$. Mass cancels — $v_\max$ is independent of vehicle mass. When $\mu_s \approx 0$ (ice), $v_\max \approx 0$; a banked curve always permits a higher $v_\max$ for the same $r$.

Add the highlighted flat-curve formula and the mass-independence result to your notes before the check below.

Misconceptions — final check

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