More practice questions

4. Q4. A car of mass 1000 kg takes a level curve of radius 50 m at 15 m s^⁻¹. (a) What friction force does the turn need? (b) If μ_s = 0.4, can the car make the turn? Take g = 9.8 m s⁻².

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   (a) The friction must supply the centripetal force:

   $$f = \frac{mv^{2}}{R} = \frac{1000 \times 225}{50}$$

   which is 4500 N.

   (b) The most static friction can give is μ_smg, that is 0.4 × 9800 or 3920 N. This is less than 4500 N, so the car cannot make the turn at this speed: it skids outward.

5. Q5. A stone tied to a string is whirled in a horizontal circle. The string suddenly breaks. In which direction does the stone move just after that?

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   Along the tangent to the circle at the point where the string broke, in the direction it was moving at that instant. At the instant of release its velocity is continuous and tangential; afterwards gravity changes the velocity and curves the path downward. It does not fly radially outward. Compare NCERT Exercise 4.22.

6. Q6. Why is the outer rail of a curved railway track laid higher than the inner rail?

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   Raising the outer rail banks the track. The push of the rails on the wheels is then tilted towards the centre of the curve, so its horizontal part supplies most of the centripetal force the train needs. Without banking, the whole centripetal force would have to come from the sideways push of the outer rail on the wheel flanges, which wears both and makes derailment more likely.

7. Q7. A car of mass 1200 kg goes round a curve banked at 10° at exactly the optimum speed. Find the normal force on it and compare it with its weight. Take g = 9.8 m s⁻².

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   At the optimum speed no friction acts, and vertically N cos θ = mg:

   $$\begin{aligned} N &= \frac{mg}{\cos 10^\circ} = \frac{11\,760}{0.9848} \\ &\approx 1.19 \times 10^{4}\ \text{N} \end{aligned}$$

   The weight is 11 760 N, so the normal force is about 1.5% larger than the weight, not smaller as on a slope.

8. Q8. A cyclist rides at 18 km/h round a level turn of radius 10 m. The coefficient of static friction between the tyres and the road is 0.3. Will the cyclist slip? Take g = 9.8 m s⁻².

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   18 km/h is 5 m s^⁻¹, so v^² is 25 m^² s^⁻². The road can supply enough friction if v^² ≤ μ_sRg, and μ_sRg is 0.3 × 10 × 9.8, that is 29.4 m^² s^⁻². Since 25 is less than 29.4, the cyclist does not slip.