Circular Motion: Numerical Problems Set (2)

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Tags Circular Motion

Question

A car of mass 1000 kg1000\text{ kg} negotiates a circular curve of radius 50 m50\text{ m} at a speed of 20 m/s20\text{ m/s} on a level road. (a) Find the centripetal acceleration. (b) Find the minimum coefficient of friction needed between tyres and road. (c) If the road is banked at angle θ\theta for this speed (no friction), find θ\theta. Take g=10 m/s2g = 10\text{ m/s}^2.

Solution — Step by Step

ac=v2r=(20)250=40050=8 m/s2a_c = \frac{v^2}{r} = \frac{(20)^2}{50} = \frac{400}{50} = 8\text{ m/s}^2

This is directed towards the centre of the curve.

On a flat road, friction provides the entire centripetal force.

μmg≥mv2r  ⟹  μmin⁡=v2rg=40050×10=0.8\mu mg \geq \frac{mv^2}{r} \implies \mu_{\min} = \frac{v^2}{rg} = \frac{400}{50 \times 10} = 0.8

When the road is banked, the horizontal component of the normal force supplies the centripetal force.

tan⁡θ=v2rg=400500=0.8\tan\theta = \frac{v^2}{rg} = \frac{400}{500} = 0.8 θ=arctan⁡(0.8)≈38.7°\theta = \arctan(0.8) \approx 38.7°

ac=8 m/s2a_c = 8\text{ m/s}^2, μmin⁡=0.8\mu_{\min} = 0.8, θ≈38.7°\theta \approx 38.7°.

Why This Works

For circular motion, some force must point towards the centre. On a flat road, only friction can do this — that is why bald tyres on rainy roads send cars sliding outward.

Banking does the same thing geometrically: tilt the surface, and a component of the normal force (which is large) takes over the role friction was playing.

Alternative Method

In the rotating frame of the car, a pseudo-force mv2/rmv^2/r pushes outward. For equilibrium, friction or banking must cancel it. Setting up the force balance in this frame gives the same equations — useful for non-inertial-frame problems in JEE Advanced.

Common Mistake

For the banking angle, students often write sin⁡θ=v2/(rg)\sin\theta = v^2/(rg). The correct relation involves tan⁡θ\tan\theta. The error comes from not resolving along the inclined surface vs. horizontally — always draw the FBD in horizontal/vertical axes, not along the slope.

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