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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Chavi on June 05, 2010, 07:05:53 pm

Title: Banked circular motion with friction
Post by: Chavi on June 05, 2010, 07:05:53 pm: Chances of being the exam?
Anyone have any easy formulas for such a situation?
Title: Re: Banked circular motion with friction
Post by: Stroodle on June 05, 2010, 07:11:38 pm: I reckon there's a pretty good chance one will be on the exam.

I always just use:

$NR\times cos(\theta)=mg$

$NR\times sin(\theta)=\frac{mv^2}{r}$

$\Rightarrow tan(\theta})=\frac{v^2}{rg}$
Title: Re: Banked circular motion with friction
Post by: Chavi on June 05, 2010, 07:21:35 pm: What about the friction?
Title: Re: Banked circular motion with friction
Post by: Greggler on June 05, 2010, 07:27:28 pm: Do you have any examples of a question involving friction + banking tracks? I've never seen any...

the most complicated questions involving friction ive encountered are towing on an inclined plane.
And banked tracks can be easily dealt with, with he equations stroodle has shown.
Title: Re: Banked circular motion with friction
Post by: Stroodle on June 05, 2010, 07:30:06 pm: $Friction+mgsin(\theta)=\frac{mv^2}{r}cos(\theta)$
Title: Re: Banked circular motion with friction
Post by: ks04 on June 05, 2010, 09:42:23 pm: Quote from: Stroodle on June 05, 2010, 07:11:38 pm
I reckon there's a pretty good chance one will be on the exam.

I always just use:

$NR\times cos(\theta)=mg$

$NR\times sin(\theta)=\frac{mv^2}{rg}$

$\Rightarrow tan(\theta})=\frac{v^2}{rg}$

Would you mind explaining these in words - It's the first time I've come accross these formulas.
Title: Re: Banked circular motion with friction
Post by: Stroodle on June 05, 2010, 10:09:28 pm: Opps. I accidentally copied one of them from my cheat sheet wrongly. Fixed now though... Sorry.

Kinda hard to explain without a diagram, so here you go:
(http://img39.imageshack.us/img39/4108/39423390.jpg)
Title: Re: Banked circular motion with friction
Post by: ks04 on June 06, 2010, 12:16:26 pm: Silly question... F_n is the Normal reaction force? which is = to m.g/cosθ?
Title: Re: Banked circular motion with friction
Post by: Stroodle on June 06, 2010, 02:49:31 pm: Yep :)
Title: Re: Banked circular motion with friction
Post by: Elnino_Gerrard on June 06, 2010, 04:40:21 pm: Quote from: Stroodle on June 06, 2010, 02:49:31 pm
Yep :)
my tacher says Frictioncostheta + Nsintheta =mv^2/r is that correct?
Title: Re: Banked circular motion with friction
Post by: Greggler on June 06, 2010, 05:00:19 pm: For banked track questions i just draw that triangle with the Normal Force as the hypotenuse, weight force as vertical and the centripital force as the horizontal.

The you can do all your calculations from that.
Title: Re: Banked circular motion with friction
Post by: kenhung123 on June 07, 2010, 01:20:58 pm: Apologies...I thought NR meant normal*radius..
Title: Re: Banked circular motion with friction
Post by: Keyzer on June 07, 2010, 04:41:41 pm: For banked questions with fiction, you need to split the components of the normal force and the friction force acting down the slope. These components will equal your Net Force.
Title: Re: Banked circular motion with friction
Post by: TyErd on June 07, 2010, 04:49:21 pm: can you explain that further Keyzer? perhaps with an example :)
Title: Re: Banked circular motion with friction
Post by: Keyzer on June 07, 2010, 09:20:46 pm: Quote from: TyErd on June 07, 2010, 04:49:21 pm
can you explain that further Keyzer? perhaps with an example :)

Well I can't upload a picture because my net is lagging like hell. But anyway, say you have a car on a banked track, 30 degree banking, with a frictional force of say 12000 N acting. You split the car into its forces, you'll have the weight force going straight down, and a normal force perpendicular to the plane. You also have the friction force going down into the bottom of the bank. To find the net force, you need to find all of the components of the forces which act in towards the centre of motion. In this case, you have the horizontal component of the normal force, N sin 30, and the horizontal component of the frictional force, which is 12,000 cos 30. These two components equal the net force, which is basically (mv^2)/R.

Try to draw this if you don't understand.
Title: Re: Banked circular motion with friction
Post by: cameron_15 on June 07, 2010, 09:24:17 pm: It makes sense when you draw triangles... Like most physics :2funny: