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.
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.
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.
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.)
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.
No big difference, it's just that your formula works when we consider that there's no friction (μ = 0), if you try to substitute μ = 0 into the formula, you get v² = gR(sinθ)/cosθ
If theta is the unknown, you often end up plugging in all of the other known values and solving for (sin theta / cos theta = some stuff) which is tan theta. = some stuff. Then, you can typically solve for theta using inverse tangent theta = that stuff.
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.
The best explanation on utube.Thank you very much
I appreciate this. I hope it was useful!
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. 👍
7:51 for those of us who just want the answer. I knew the math would be above my head. Great video!
Thanks for the feedback and it's a good call to post in the caption where the final answers are. I'll fix that now.
thank you so much, i literally couldnt find anyone else doing this type of problem
You're welcome! I'm happy it could help. 👍
you just saved my physics grade. 100% soooonnnnnnnn
Let's goooooo
great video man!!!
I appreciate this. 👍
You're the best man.
YOU'RE the best. Hope it helps you ace the exam.
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!
Better explained🎉🎉🎉
5:30 horizontal formula: Must be -mu*N*cos(teta), why you did positive not negative sir?
if accelaration towards center the friction must be opposite direction?
A component of friction and a component of normal force point toward the center. There are no forces with components toward the outside of the curve.
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.
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.
No big difference, it's just that your formula works when we consider that there's no friction (μ = 0), if you try to substitute μ = 0 into the formula, you get v² = gR(sinθ)/cosθ
How would you solve for theta here?
If theta is the unknown, you often end up plugging in all of the other known values and solving for (sin theta / cos theta = some stuff) which is tan theta. = some stuff. Then, you can typically solve for theta using inverse tangent theta = that stuff.
youtube.com/@gravitation_gravitivity?si=S1hTzhCgctn2EX4v
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You can do it! If this stuff makes people sad, it is usually the trigonometry. Maybe this can help? th-cam.com/video/Lpm2-pWpUhs/w-d-xo.html
@@dr.piercesphysicsmath9071 thank you sir❣️