Centripetal force mcq

The centripetal force is the resultant force equal to mv^2/r. The force exerted on the car by the bridge is the reaction force which must be less than the weight of the car so that the resultant force can act downwards. This means that mv^2/r = mg - R where R is the reaction force so rearranging that you get R = mg - mv^2/r. I hope that made sense.

Yeah that makes sense thank you
Suppose this road was a full circle and the car could travel to the side and upside down too (like a roller coaster) what would the force of the road on the car be then when it's on the right or left side of the circle?

I'm not sure how it would work in that situation as neither the reaction force or weight would be acting towards the centre of the circle. Possibly a component of friction? However a similar(ish) example would be clothes in a washing machine. When they're on the sides the force on the clothes (the reaction force) would be equal to the centripetal force mv^2/r.

Ok thank you sm!
If it isn't too much trouble can you please tell me what the forces act sideways when we have a roller coaster in an outward loop?

I believe the following link would answer most of your questions around circular motion.
https://ux1.eiu.edu/~cfadd/1350/Hmwk/Ch06/Ch6.html

Why is D??
Why not B?

Put in the equals: F = ma = mg = mv^2/r , hence a = g = v^2/r.
mg and mv^2/r are not separate entities your just describing/resolving the effect of gravitational force which is causing centripetal motion.

Yes
Thank you
Do you have any idea what would be the NRF for a roller coster when it's sideways on a loop?

I don't even know what NRF means: normal resistance force? i have no information about the question at hand here

Yes normal reaction force

Ok thank you sm!
If it isn't too much trouble can you please tell me what the forces act sideways when we have a roller coaster in an outward loop?

well it would be quite hard to say just off the info you've provided.

But draw a free body diagram, the Normal force acts perpendicular to the interacting surface and always opposes the velocity vector and is normally just the opposing magnitude force multiplied by mu (the static friction coefficient). But if you draw the FBD correctly then annotate the resolving forces it should be more apparent.

I don't know about that situation sorry. I don't think you'll need to know about that for A level (I'm assuming) physics though.

Centripetal formula gives resultant force -always acts to centre of circle
you know then mv^2/r= force towards centre minus force out (as it is resultant force towards centre)
so mv^2/r =mg-R