Normal reaction in banked road circular motion problem

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Main Question or Discussion Point

What is the equation of forces please for the axis parallel to the banked road? Is it R1+R2-mgcosθ=mV^2/r or R1+R2-mgcosθ=-mV^2/r? Can you advise me please?
 

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  • #2
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What is the equation of forces please for the axis parallel to the banked road? Is it R1+R2-mgcosθ=mV^2/r or R1+R2-mgcosθ=-mV^2/r? Can you advise me please?
The friction will be directly opposite mgsinθ if the car is going slow enough, but it will be the other way if the object is going fast enough. Therefore, the equation depends on which way the car is slipping. If it is in perfect equilibrium, then mgsinθ and friction cancel out and you're left with only components perpendicular to the road (such as the normal force and mgcosθ) causing the centripetal acceleration.
 
  • #3
sophiecentaur
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What is the equation of forces please for the axis parallel to the banked road? Is it R1+R2-mgcosθ=mV^2/r or R1+R2-mgcosθ=-mV^2/r? Can you advise me please?
If you want an answer to that then it is probably better if you actually define all the variables in those equations, along with a suitable diagram. Also, you will get a better response if you can manage to be a bit more polite with your replies; the customer is not always right here because there are no (paying) customers. "That's all".
 
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R1 and R2 are the normal reactions of the ground to the vehicle's wheels
mg is the vehicle weight
V it's speed
θ the angle of the banked road
I think now it's clear
 
  • #5
jbriggs444
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R1 and R2 are the normal reactions of the ground to the vehicle's wheels
mg is the vehicle weight
V it's speed
θ the angle of the banked road
Is the angle measured from the horizontal or from the vertical? Is positive for R1, R2 and the centripetal acceleration taken in the sense of an inward net force? I assume yes. Is positive for mg cos(θ) taken in terms of down=positive or up=positive. That would obviously have a key influence on the answer you seek.

Since you have adopted coordinates in which the centripetal acceleration is neither aligned with the x axis or with the y axis and have restricted your attention to the x axis, one would expect only part of the centripetal force (##\frac{mv^2}{r}##) to be relevant. Yet there is no cos(θ) or sin(θ) factor on the term to account for this. Similarly, there is no cos(θ) or sin(θ) factor on the terms for R1 or R2. So I have to question whether either equation can be correct.

Still, a simple argument can be made. If we select a track with a degenerate bank angle so that cos(θ) = 0 then the equations reduce to

##R_1 + R_2 = \frac{mv^2}{r}## and
##R_1 + R_2 = -\frac{mv^2}{r}##

Assuming a convention that positive is inward/up then ##R_1## and ##R_2## must be positive. ##\frac{mv^2}{r}## will neccessarily be positive. So the second equation is obviously wrong. Though, as above, that does not mean that the first equation must be right.
 

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