Circular Motion: PYQ Walkthrough (5)

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Tags Circular Motion

Question

A car of mass m=1000m = 1000 kg takes a turn of radius r=50r = 50 m on a level road. The coefficient of static friction between the tyres and road is μs=0.5\mu_s = 0.5. Find the maximum safe speed of the car. If the road is banked at angle θ=30°\theta = 30°, find the new maximum speed (still with the same μs\mu_s). (JEE Main 2024 Shift 1)

Solution — Step by Step

For maximum speed, μsmg=mv2r\mu_s m g = \frac{m v^2}{r}. So:

vmax⁡=μsgr=0.5×10×50=250≈15.8 m/sv_{\max} = \sqrt{\mu_s g r} = \sqrt{0.5 \times 10 \times 50} = \sqrt{250} \approx 15.8 \text{ m/s}

For a banked road with friction, max speed is:

vmax⁡=rg(μs+tan⁡θ)1−μstan⁡θv_{\max} = \sqrt{\frac{r g (\mu_s + \tan\theta)}{1 - \mu_s \tan\theta}}

tan⁡30°=1/3≈0.577\tan 30° = 1/\sqrt{3} \approx 0.577.

Numerator: μs+tan⁡θ=0.5+0.577=1.077\mu_s + \tan\theta = 0.5 + 0.577 = 1.077.

Denominator: 1−μstan⁡θ=1−0.5×0.577=1−0.289=0.7111 - \mu_s \tan\theta = 1 - 0.5 \times 0.577 = 1 - 0.289 = 0.711.

 vmax⁡=50×10×1.0770.711=538.50.711=757.4≈27.5 m/sv_{\max} = \sqrt{\frac{50 \times 10 \times 1.077}{0.711}} = \sqrt{\frac{538.5}{0.711}} = \sqrt{757.4} \approx 27.5 \text{ m/s}

Final answers: vmax⁡,level≈15.8v_{\max,\text{level}} \approx 15.8 m/s; vmax⁡,banked≈27.5v_{\max,\text{banked}} \approx 27.5 m/s.

Why This Works

On a level road, friction is the only force pointing inward, so it provides the entire centripetal force. The bound μsmg\mu_s m g caps the centripetal force, which caps the speed.

On a banked road, the inward component of normal force also contributes to centripetal force, allowing higher speeds. Friction adds further margin. The combined formula handles both effects.

Alternative Method

Set up the inclined-plane FBD on the banked road: resolve the normal force NN and friction f=μsNf = \mu_s N along horizontal and vertical axes. Vertical: Ncos⁡θ−fsin⁡θ=mgN\cos\theta - f \sin\theta = mg. Horizontal: Nsin⁡θ+fcos⁡θ=mv2/rN\sin\theta + f\cos\theta = mv^2/r. Solve simultaneously to recover the boxed formula.

For a banked road without friction, v=rgtan⁡θv = \sqrt{rg \tan\theta}. With friction, the formula above. Memorise both — JEE often asks for both in the same question.

Common Mistake

Forgetting that on a banked road, friction can act in either direction depending on speed. For maximum speed, friction acts down the slope (inward, helping). For minimum speed, friction acts up the slope (outward).

Plugging sin⁡θ\sin\theta instead of tan⁡θ\tan\theta in the formula. The clean derivation always ends in tan⁡θ\tan\theta because we divide the two equations. Recheck if you see sin⁡θ\sin\theta alone in the final expression.

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