A horizontal curve on a racing track is banke | vedclass

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(A) For a vehicle moving on a banked road at the safe speed (where no friction is required),the forces acting on the vehicle are the gravitational for

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$g$

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(A) For a vehicle moving on a banked road at the safe speed (where no friction is required),the forces acting on the vehicle are the gravitational force $(mg)$ acting downwards and the normal force $(N)$ acting perpendicular to the banked surface.The vertical component of the normal force balances the weight: $N \cos 	heta = mg$.The horizontal component of the normal force provides the necessary centripetal force: $N \sin 	heta = m a_c$,where $a_c$ is the centripetal acceleration.Dividing the two equations: $\frac{N \sin 	heta}{N \cos 	heta} = \frac{m a_c}{mg}$.This simplifies to: $	an 	heta = \frac{a_c}{g}$.Therefore,the centripetal acceleration is $a_c = g 	an 	heta$.Given the banking angle $	heta = 45^o$,we have $a_c = g 	an(45^o) = g(1) = g$.

$A$ horizontal curve on a racing track is banked at a $45^o$ angle. When a vehicle goes around this curve at the curve's safe speed (no friction needed to stay on the track),what is its centripetal acceleration?

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