Q. 6

# A circular road of radius r is banked for a speed υ = 40 km/hr. A car of mass m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible.

A. The car cannot make a turn without skidding.

B. If the car turns at a speed less than 40 km/hr, it will silp down.

C. It the car turns at the correct speed of 40 km/hr, the force by the road on the car is equal to

D. If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than

Answer :

It is given that the road is banked for a speed of 40km/hr. If the car continues at this speed, it is possible that it turns without skidding. Hence, (A) is incorrect.

However, if it slows down than this speed, then the forces on the car in inward and outward directions will not cancel out and net force will be inwards. Therefore, it will slip down. (B.) is true.

When the car turns at the exact speed of 40km/hr,

Net force on the car by the road= mv^{2}/r(sinθ)

where θ = angle of banking

⇒ Option (C) is false.

Since, sinθ≤1

⇒ rsinθ ≤ r

⇒ mv^{2}/rsinθ ≥ mv^{2}/r

⇒ Force on car by road ≥ mv^{2}/r

We know that the normal reaction N=mg

And, N cosθ=mv^{2}cosθ/r

⇒ mg=mv^{2}cosθ/r

Also, cosθ≤1

⇒ mv^{2}cosθ ≤ mv^{2}

⇒ mv^{2}cosθ/r ≤ mv^{2}/r

⇒ mg ≤ mv^{2}/r

⇒ mg ≤ mv^{2}/r sinθ

⇒ mg ≤ Force on car by road θ

Therefore, (D.) is true.

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The displacement vector of a particle of mass m is given by r (t) = î A cos ωt + ĵ B sin ωt.

(a) Show that the trajectory is an ellipse.

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