Circular Motion: Real-World Scenarios (1)

easy • 3 min read
Tags Circular Motion

Question

A car of mass 10001000 kg negotiates a flat curve of radius 5050 m at a speed of 1515 m/s. The coefficient of static friction between the tyres and road is μs=0.5\mu_s = 0.5. Will the car skid? If not, what is the maximum safe speed on this curve?

Solution — Step by Step

On a flat (un-banked) road, only friction can provide the centripetal force pulling the car into the curve. Gravity and normal reaction are vertical and cancel; they do nothing for centripetal motion.

The required centripetal force is Fc=mv2/rF_c = mv^2/r. The maximum available friction is fmax⁡=μsmgf_{\max} = \mu_s mg. The car does not skid as long as:

mv2r≤μsmg  ⟹  v2≤μsgr\frac{mv^2}{r} \leq \mu_s mg \implies v^2 \leq \mu_s g r

Mass cancels — the result is independent of how heavy the car is.

 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}

The car is moving at 1515 m/s, which is less than 15.815.8 m/s, so it does not skid — but only just barely.

Final Answer: No skid. Maximum safe speed ≈15.8\approx 15.8 m/s (∼57\sim 57 km/h).

Why This Works

Friction is a passive force — it provides whatever centripetal force is needed up to its maximum value μsmg\mu_s mg. The moment the required centripetal force exceeds this cap, the tyres lose grip and the car slides outward (which is why skids on curves are usually outward, not inward).

Banking the road tilts the normal force to contribute toward the centre, allowing higher safe speeds without relying entirely on friction. That’s why highway curves are banked.

Alternative Method

Frame it in terms of “centripetal acceleration” instead of force: ac=v2/r=225/50=4.5a_c = v^2/r = 225/50 = 4.5 m/s2^2. Maximum frictional acceleration available =μsg=5= \mu_s g = 5 m/s2^2. Since 4.5<54.5 < 5, no skid. Same physics, slightly leaner algebra.

Including the car’s weight mgmg in the “force balance” along with mv2/rmv^2/r is wrong. Weight acts downward, centripetal force acts horizontally — they’re perpendicular, so they don’t sum. Always draw the FBD and check directions before adding forces.

For a banked road at angle θ\theta with friction, the maximum speed is vmax⁡=gr(tan⁡θ+μ)/(1−μtan⁡θ)v_{\max} = \sqrt{gr(\tan\theta + \mu)/(1 - \mu\tan\theta)}. Drill this formula — JEE Main asks it almost every alternate year.

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