What angle does a pendulum make in a car taking a turn at 54 km/h with radius 20m?

θ=tan⁻¹(1.125). Using tan θ=v²/(rg)=15²/(20×10)=225/200=1.125.

What formula gives pendulum angle in a turning car?

tan θ = v²/(rg) where v is car speed, r is turning radius, g is gravity.

Car 54 km/h takes turn of radius 20m — angle of pendulum with vertical

JEERANKERS

PhysicsMechanics

QA car moving with a speed of 54 km/h takes a turn of radius 20 m. A simple pendulum is suspended from the ceiling of the car. Determine the angle made by the string of the pendulum with the vertical during the turning. (Take $g = 10$ m/s²)

A) $\tan^{-1}(0.5)$ B) $\tan^{-1}(0.75)$ C) $\tan^{-1}(1.125)$ D) $\tan^{-1}(0.25)$Circular Motion

MCQ

✅ Correct Answer

C) tan⁻¹(1.125)

✓

Solution

Convert speed to m/s

$v=54\text{ km/h}=54\times\dfrac{1000}{3600}=15\text{ m/s}$

Identify forces on pendulum bob

In the non-inertial frame of the car, the bob experiences:

• Weight $mg$ downward
• Pseudo-force $\dfrac{mv^2}{r}$ outward (centrifugal)

Find angle of string

$\tan\theta=\dfrac{\text{centrifugal force}}{\text{gravity}}=\dfrac{mv^2/r}{mg}=\dfrac{v^2}{rg}$

$=\dfrac{(15)^2}{20\times10}=\dfrac{225}{200}=\dfrac{9}{8}=1.125$

$\theta=\tan^{-1}(1.125)$

📘 Pendulum in Non-Inertial Frame

In a car turning with centripetal acceleration $a_c=v^2/r$, the pendulum equilibrium is set by the resultant of gravity $g$ (downward) and pseudo-force $a_c$ (radially outward). The string aligns with the resultant: $\tan\theta=a_c/g=v^2/(rg)$.

💡

Important Concepts

Pseudo-Force in Rotating FrameIn the non-inertial frame of the car, a centrifugal pseudo-force of magnitude mv²/r acts outward. The pendulum bob reaches equilibrium when the string tension balances both gravity and this pseudo-force.

tan θ = v²/rgThis is the standard formula for the angle of a pendulum in a horizontally accelerating system. Here v=15m/s, r=20m, g=10: tan θ=225/200=1.125.

Unit Conversion Key54 km/h × (1000m/km)/(3600s/h) = 15 m/s. Always convert to SI before substituting.

Physical Meaningθ≈48.5° from vertical — the pendulum swings significantly outward during the turn, consistent with the high centripetal acceleration (11.25 m/s² vs g=10 m/s²).

FAQs

Why use tan θ = v²/rg?

⌄

In the car's frame, the bob is in equilibrium under tension T, weight mg downward, and pseudo-force mv²/r outward. T·cos θ=mg and T·sin θ=mv²/r. Dividing: tan θ=v²/rg.

What is the centripetal acceleration?

⌄

a_c = v²/r = 225/20 = 11.25 m/s². This exceeds g=10 m/s², so θ>45°.

What if r were much larger?

⌄

tan θ = v²/rg → 0 as r → ∞. The pendulum barely deflects on gentle curves.

Can we solve in the ground frame?

⌄

Yes: The pendulum is moving in a circle with centripetal acceleration v²/r directed inward. Net inward force = T·sin θ = mv²/r. Upward: T·cos θ=mg. Same result.

Is this from JEE Main 2026?

⌄

Yes, this appeared in JEE Main 2026.

🔗