Banked Turn with Friction - Physics of Speed Limits on Banked Curves
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- Опубліковано 13 чер 2022
- We take a look at the general case of finding the maximum speed at which a car can drive around a banked curve without skidding out. 0:00 We set up the problem using F = ma (Newton's 2nd Law) in the vertical (a = 0) and horizontal (a = mv^2 / R) directions to 4:17 relate the maximum speed to the ramp angle, radius of curvature of the road, and the coefficient of friction. Finding the speed limit on a banked curve is a very standard problem in mechanics in undergraduate physics, AP Physics, and IB Physics, and is useful for those who design banked roads. 7:48 Finally, we show the specific results for maximum speed vs. ramp angle for the specific case of an exit ramp radius of curvature of 100 m, acceleration due to gravity of 10 m/s^2, and a coefficient of friction of 0.9, which is reasonable for a dry day for rubber on asphalt. In addition, we explore what would happen if the road were unbanked and if the road were banked (tilted) the wrong way.
just want to say you're a lifesaver - really appreciate how visual your diagrams are, finally got my head around it :)
Oh, I love to hear that! Thanks for taking the time to let me know. 👍
Great video. I have always found banked curves on roads and highways to be aesthetically pleasing.
Beyond the critical angle (48 deg in the example), the car cannot skid off the track; it will stay on the track even with speed set to infinity. In practice though, a car can only take so much g-force.
thank you so much, i literally couldnt find anyone else doing this type of problem
You're welcome! I'm happy it could help. 👍
Thanks. I'm in a horribly run physics course at the moment, and this really helped me to understand the problem.
You are welcome! And you are doing the right thing by taking the initiative to find resources to build the ideas yourself. This can be interesting stuff; independent of the teacher and course. Good luck with the course. 👍
Better explained🎉🎉🎉
you just saved my physics grade. 100% soooonnnnnnnn
Let's goooooo
thanks a lot. finally makes sense. i guess my way of doing FBD where gravity is off to the side and everything else on the axis's doesn't work for this kind of problem although makes other problems a lot more convenient.
Your way does actually work, but you have to also do trig on the "a" side of f=ma since a would not be along either of your tilted axes.
tried it for fun but it gets pretty ugly as you cant set Fnety to 0 and end up with a system of equations with 2 unknowns a and N. tried solving for each but came up with funny results.
Brilliant video. What equation did you use to get that graph? I don't know if I'm being completely stupid but I gave a graphing calculator the equation shown with the numbers given (R = 100, etc.) and got a totally different tan graph, and now I'm confused. Also, I used this process to find the max speed with different dimensions (R = 15, theta = 75, and COF = 0.32). Putting this in my calculator gives me a math error, as the denominator ends up negative. What does that mean in the context of turning? Are those just unfeasible dimensions? Hope you can help clear this up.
Hmm... I did just use the equation shown and it should work for your parameters, even though they'd be extreme for a road. Maybe a radian mode / degree mode issue? Or one little missing set of parentheses? Those are the usual culprits with my students.
@@dr.piercesphysicsmath9071 Problem solved, desmos was in radians. Thanks!
I am confused, what is the difference between v²=R.g.tan theta and the above equation?
The equation you give seems to be for a different problem that does not involve a friction coefficient. Likely, your equation is derived from F=ma, but for a different scenario than the one considered here.
what about mg sin theta when adding the sum of forces in the x direction (the car is on an incline isn't it?).
Also how would you do this question that says: "A car can barely negotiate a 50m unbanked curve when the coefficient of static friction between the tires and road is 0.80. How much bank would the curve require if the car is to safely go around the curve without relying on friction?"
I did not tilt my axes along the ramp. So, I didn't need to do any trigonometry with mg. I did need to do trig with normal force and friction, though. If you do tilt your axes along the ramp (this is another way to solve), you will have to do trig with acceleration and also with mg.
First problem: make axis toward center. F = ma horizontally would give umg = mv^2 / R. So, v = root (Rug).
Second problem: make an axis horizontally toward the center. Then horizontal component of friction and horizontal component of normal force = mv^2 / R.
Vertically, you'd have no acceleration, so no net vertical force. So, vertical part of normal force = weight + vertical component of friction. (Upward force = sum of downward forces.)
why isnt the value of N equal to cos(theta)*mg? isnt it usually the case when objects are on a curb?
The car is accelerating toward the center, so N is LARGER than mg cos theta. N helps to provide some of the "extra" force needed to help the car turn.
😢😢😢
You can do it! If this stuff makes people sad, it is usually the trigonometry. Maybe this can help? ua-cam.com/video/Lpm2-pWpUhs/v-deo.html
@@dr.piercesphysicsmath9071 thank you sir❣️