Car, banked road, design speed question. Help please :)

[PB]

Curves of less than 400m radius should be avoided on roads when a car is travelling at 100km/hr.
Calculate the banked angle of this road for the above design speed.
Thanks, +1

[availn]

I'm assuming this is a frictionless road? In which case, the centripetal acceleration needed is:

v²/r = (27.78)²/400 = 1.929ms^-2

This acceleration is provided wholly by the component of gravitational acceleration down the slope. So:

g·sin(θ) = 1.929
∴ θ = 11.12°

[Alwin]

A "short-cut" method for these qs

Assuming frictionless as well:


@avalin, you might want to check your answer,

[PB]

Yep Yep, thanks Alwin! Thank you also availn - you got close
@Alwin, I am guessing that this tan(thet)=v(sq)/gr is a known formula because thats what the solns used - but I was like -whaaat? cus i have never seen it before - not even in the textbook.
I am not sure how this formula is derived 8O but ah well, as long as it works for these design speed questions .

Do you know where it comes from? It seems to be a pretty random formula...I assume it comes from the stadard design speed formula v=sqrt(gr)?

[psyxwar]

A "short-cut" method for these qs

loving the smiley face!

[Alwin]

Yep Yep, thanks Alwin! Thank you also availn - you got close
@Alwin, I am guessing that this tan(thet)=v(sq)/gr is a known formula because thats what the solns used - but I was like -whaaat? cus i have never seen it before - not even in the textbook.
I am not sure how this formula is derived 8O but ah well, as long as it works for these design speed questions .

Do you know where it comes from? It seems to be a pretty random formula...I assume it comes from the stadard design speed formula v=sqrt(gr)?

If we look at my pic from the other post:
Vertically:


Horizontally:


If we divide the two equations (or you can solve by other means if you want )

By simplifying, we get:


Hope it makes sense

EDIT: and thankyou METAPOD/Psyxwar

EDIT 2: slipped up in a few places, ty guys + if there's STILL something wrong, please point it out

[Chazef]

alwin I think you mean tan theta

[Alwin]

alwin I think you mean tan theta

THANKS!! im always making stupid mistakes like that.. esp when I used it correctly in the previous post.

Fixed it up, ty Chazef

[PB]

Cheers Alwin btw, I think you should swap horizontal and vertical in your explanation!

[BasicAcid]

If we look at my pic from the other post:
Vertically:


Horizontally:


If we divide the two equations (or you can solve by other means if you want )

By simplifying, we get:


Hope it makes sense

EDIT: and thankyou METAPOD/Psyxwar

EDIT 2: slipped up in a few places, ty guys + if there's STILL something wrong, please point it out

Also Alwin, did they teach you this at school or did you have a tutor?
Or did you just read up about this stuff yourself?
If so, do you have any sources I could look at? I find these connections really interesting.

By the way, do you have the cheat sheets you used last year/a compilation of these shortcut formulas I could have a read of? I'm just wondering if I'm missing out on any haha.

Thank you.

[Alwin]

Also Alwin, did they teach you this at school or did you have a tutor?
Or did you just read up about this stuff yourself?
If so, do you have any sources I could look at? I find these connections really interesting.

By the way, do you have the cheat sheets you used last year/a compilation of these shortcut formulas I could have a read of? I'm just wondering if I'm missing out on any haha.

Thank you.

Nah, never been to tutor for anything.  I prefer to just derive the formulas myself, the maths in VCE physics is pretty easy.

Didn't make a "proper" cheat sheet (same for methods if you're wondering), just some basic formulas because some "shortcuts" only apply in certain situations. Normally I'd just come across one in the textbook solution or practise exam solutions and then go about proving them myself.
You can tackle all the problems in vce physics without the need for these *special formulas*

[lzxnl]

Also Alwin, did they teach you this at school or did you have a tutor?
Or did you just read up about this stuff yourself?
If so, do you have any sources I could look at? I find these connections really interesting.

By the way, do you have the cheat sheets you used last year/a compilation of these shortcut formulas I could have a read of? I'm just wondering if I'm missing out on any haha.

Thank you.

It's not THAT complicated...it's just resolving vertical and horizontal components; you know that the car only moves horizontally, so the vertical component of the net force is zero. Resolve the forces to zero.
Then you know that the net force is mv^2/r horizontally...so resolve the forces to that. You could read this in a textbook...

NOTE I have to add something. Here, N cos theta = mg. Normally you have N = mg cos theta on an inclined plane.
Why the discrepancy? Well, the net force in the direction of the normal to the surface is NOT zero in the banked curve motion but it is for the inclined plane. Just thought I'd clarify this.

[Alwin]

NOTE I have to add something. Here, N cos theta = mg. Normally you have N = mg cos theta on an inclined plane.
Why the discrepancy? Well, the net force in the direction of the normal to the surface is NOT zero in the banked curve motion but it is for the inclined plane. Just thought I'd clarify this.

Uhh.. NOTE that I have to add something here:
The discrepancy is because on an inclined plane, the direction of the net force is parallel to the slope:

diagram

Therefore:

On a banked track, we take the direction of the net force as horizontal to the ground:

diagram

Therefore: REGARDLESS IF THE NET FORCE IS ZERO OR OTHERWISE

It's just how we chose to resolve the forces

[PB]

uh ok you two lost me but its alrite, I got the formula - im not confused
@basicacid every textbook has its weaknesses - thats why you read from different sources! But i don't think even jacaranda had it.

[lzxnl]

Uhh.. NOTE that I have to add something here:
The discrepancy is because on an inclined plane, the direction of the net force is parallel to the slope:

diagram

Therefore:

On a banked track, we take the direction of the net force as horizontal to the ground:

diagram

Therefore: REGARDLESS IF THE NET FORCE IS ZERO OR OTHERWISE

It's just how we chose to resolve the forces

Our explanations are the same. If the direction of the net force is parallel to the slope, there is NO force in the direction of the normal.