0
$\begingroup$

I am currently working on the banked turn problem. I know that once you set up all the equations properly you end up with $$v^2=gR\tan{\theta}$$

I drew a little diagramm of the forces:

enter image description here

I know that the right way to do it is to separate the Normal force $N$ into the two lighblue components $\color{blue}{N\sin{\theta}=\frac{mv^2}{R}}$ $\color{blue}{N\cos{\theta}=mg}$ and after dividing the equations I arrive at the relation $v^2=gR \tan{\theta}$

My question is this: Why do we separate into the blue components and not into something containing the red component. In other words, isn't the red component the one pointing in the direction of the centripetal force $\frac{mv^2}{R}$?

$\endgroup$
0
$\begingroup$

The red line isn't really pointing in the direction of the centripetal force.

Consider the top-down view of the situation, as compared to the side-view we have now. To maintain a curvature, the vector has to point inwards, which would correspond to the horizontal blue line.

The red line you call out would be pointing both towards the centre of the circle, and downwards. This downwards component force wouldn't be contributing to the circular motion, therefore the red vector would not be the centripetal force.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your answer. If I understand what you said correctly then if I were to separate N into components containing the red vector then I couldn't have said $\color{red}{\text{Force component in }x \space \text{direction}}=\frac{mv^2}{r}$ because the red vector actually is not entirely in the direction of the centripetal force right? $\endgroup$ – Nullspace Nov 15 '18 at 14:13
  • $\begingroup$ @Nullspace Yeah, exactly. I think you may have been trying to apply the centripetal force to directly oppose the inertial pseudo forces acting on it; but part of the inertia goes into opposing gravity, and that component isn't related to the centripetal force. $\endgroup$ – JMac Nov 15 '18 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.