Physics homework help - centripetal acceleration?

Question

A curve of 220m radius is banked at 18 degrees. At what speed must an automobile go around this curve if no frictional forces are to be used to keep the automobile on its circular path?

Please help me set up the problem. I need the formula, not the answer.

Answer 1

I'll take you at your word, namely that you want to know HOW TO SET THE PROBLEM UP, not just be given the answer.

(Unfortunately, that latter is what the previous responder gave you, formally, perhaps because you wrote "I need the formula." Of course, giving you the final formula DOESN'T tell you how to set the problem up. So, you really want to avoid sending mixed signals!)

If the vehicle is to round the curve without calling on any friction, then there are ONLY two forces to which it is subject:

1. Its weight, mg, acting VERTICALLY DOWNWARDS, and

2. The NORMAL reaction from the road, N, acting PERPENDICULAR to the sloped surface.

Now the centripetal acceleration is itself HORIZONTAL, and is a vector pointing towards the centre of the curve. (You probably can't imagine how much it really PLEASES me to see "centripetal acceleration" rather than "centrifugal force" in a question like this. Newton himself would have been pleased with your teacher and/or book! Did you know that Newton didn't really understand how to get his law of gravity until he had sorted out the muddle in his mind between "centrifugal force" and "centripetal acceleration"?!)

O.K., so:

A. Vertically, a component of N has to balance the weight of the vehicle. That gives you an certain (N, theta, mg) relationship, where theta = 18 degrees.

B. Horizontally, only another component of N can provide the necessary (horizontal) centripetal acceleration. That gives you an (N, theta, v^2 / r) relationship.

C. Eliminating N between the relationships A. and B. will produce the final equation, which I'm afraid you've already been given, and thereby determine for you what ' v ' must be. But at least you will now have the experience and pleasure of seeing how it comes about for yourself.

This was in fact a nice problem; it took me back many years, to think about how to do it. The things that help it separate into two simple initial steps are of course that (i) gravity and the centripetal acceleration are perpendicular vectors, and also (ii) with friction absent from the problem, it ONLY involves three vectors, two of them forces and the other the induced acceleration. VERY NICE!

Live long and prosper.

Answer 2

Try ac=v^2/R
where a is acceleration of circular motion is equal to the velocity squared divided by the radius.

Answer 3

In the first question the frictional force is the centripetal force. In the second question the upward frictional force balances the downward weight, so the person does not slide down. In that problem the centripetal force is the normal force caused by the wall he is pinned against.

Answer 4

Try this one tg 18º = v2/gr

tangent of 18º equals velocity square divided by (gravity times radius) so:


v= square root of ( tangent of 18º times gravity times radius)