A car is moving on an overbridge of radius R | vedclass

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(B) Let $	heta$ be the angle that the radius vector to the car makes with the vertical at any point on the overbridge.The forces acting on the car ar

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decreases

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(B) Let $	heta$ be the angle that the radius vector to the car makes with the vertical at any point on the overbridge.The forces acting on the car are its weight $mg$ (downwards) and the normal reaction $N$ (upwards).The net centripetal force required for circular motion is provided by the component of weight towards the center and the normal reaction.The equation of motion is: $mg \cos 	heta - N = \frac{mv^2}{R}$.Rearranging for the normal reaction: $N = mg \cos 	heta - \frac{mv^2}{R}$.As the car descends from point $B$ (the highest point) to point $C$,the angle $	heta$ increases from $0^\circ$ to $90^\circ$.As $	heta$ increases,$\cos 	heta$ decreases.Since $N = mg \cos 	heta - \frac{mv^2}{R}$,as $\cos 	heta$ decreases,the normal reaction $N$ also decreases.

$A$ car is moving on an overbridge of radius $R$ with a constant speed $v$. As the car is descending on the overbridge from point $B$ to $C$,the normal reaction on it due to the bridge:

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