The smell of burned rubber

From the forces acting on the vehicle in a banked curve where N = Normal from the ground ; F = Frictional force; $N\cos\theta-F\sin\theta=mg$ $N\sin \theta +F\cos\theta=\frac{ {m \ v_{b}^2}}{r}$ $F = μN$ Solving for v we get $v_{b}= \sqrt{\frac{ rg(tanθ+μ)}{(1-μtanθ)}}$ For a non-banked (flat) track of the same radius we get $v_{f}= \sqrt {rg}$ The ratio $\frac{ { v_{b}}}{v_{f}}= \sqrt{\frac{ (tanθ+μ)}{(1-μtanθ)}}$ According to the question $\theta=20,~μ=1$ So ,we get $\frac{ { v_{b}}}{v_{f}}= 1.464$

Friends

We will solve this by drawing a free body diagram(FBD) of the car on a banked as well as a flat road.

To start with...

Forces acting on the car are-

1. Weight of the car = mg (downward)
2. Centrifugal force = $\frac{mv^{2}}{r}$ (outward)
3. Force of friction = $\mu$ N (inward)

where, m=mass of car and N=normal reaction on car

Friction force is inwards because the car has a tendency to slide outwards under the action of the centrifugal force.To oppose this motion, friction force is inwards.

The FBD of car is-

Alt

The figure is self-explanatory.

Now we proceed further...

For banked road

Summation of forces parallel to incline = 0

This gives N=mgcos $\theta$ + $\frac{mv_b^{2}}{r}$ sin $\theta$

Summation of forces perpendicular to incline = 0

This gives $\frac{mv_b^{2}}{r}$ cos $\theta$ =mgsin $\theta$ + $\mu$ N

Putting value of N from above equation in this, we get

$\frac{mv_b^{2}}{r}$ cos $\theta$ =mgsin $\theta$ + $\mu$ mgcos $\theta$ + $\mu$ $\frac{mv_b^{2}}{r}$ sin $\theta$

Cancelling out 'm' on both sides, we get

$\frac{v_b^{2}}{r}$ cos $\theta$ =gsin $\theta$ + $\mu$ gcos $\theta$ + $\mu$ $\frac{v_b^{2}}{r}$ sin $\theta$

=> $\frac{v_b^{2}}{r}$ (cos $\theta$ - $\mu$ sin $\theta$ )=g(sin $\theta$ + $\mu$ cos $\theta$ ) ----- $\boxed{i}$

For flat road

Summation of forces along horizontal road = 0

This gives $\frac{mv_f^{2}}{r}$ = $\mu$ mg

=> $\frac{v_f^{2}}{r}$ = $\mu$ g (By cancelling 'm' on both sides) ----- $\boxed{ii}$

Putting value of ' $\mu$ g' from eqn-ii in eqn-i, we get

$\mu$ $\frac{v_b^{2}}{r}$ (cos $\theta$ - $\mu$ sin $\theta$ )= $\frac{v_f^{2}}{r}$ (sin $\theta$ + $\mu$ cos $\theta$ )

=> $\frac{v_b^{2}}{v_f^{2}}$ =(sin $\theta$ + $\mu$ cos $\theta$ )/ $\mu$ (cos $\theta$ - $\mu$ sin $\theta$ )=(sin $\theta$ +cos $\theta$ )/(cos $\theta$ -sin $\theta$ ) (because it is given that $\mu$ =1)

Now put value of $\theta$ ,i.e, $\theta$ =20 (Given)

We get,

$\frac{v_b}{v_f}$ =1.464 $\boxed{Ans}$

The following claim is proved first.

The maximum speed of a car which travels on a banked track with angle of inclination A and radius r is given by sqrt(rg(sinA + UcosA)/(cosA-UsinA)), where U is the friction of static co-efficient between the tyre and the track.

Proof:- Let v be the maximum velocity and the normal force acting on the car be R. The forces acting on the car in the horizontal direction are:- i) Centrifgual force (mv^2/r) outwards ii) Rsin(A) inwards iii) URcos(A) inwards.

From Newton's 2nd law of motion, mv^2/r= Rsin(A) + URcos(A).....(1)

Now consider the vertical forces acting on the car. They are:- i) Rcos(A) upwards ii) URsin(A) downwards iii) mg downwards (where mg is the weight of the car)

From Newton's 2nd law of motion, Rcos(A)= Usin(A) + mg .....(2)

Solving equations (1) and (2) for v gives us v= sqrt(rg(sinA + UcosA)/(cosA-UsinA))

Here U=1, A= 20 degree.

So, vb= sqrt(rg(sin(20 deg)+ cos(20 deg))/(cos(20 deg)-sin(20 deg))

On a flat track, A= 0

So, vf= sqrt(rg(sin(0)+ cos(0)/(cos(0)-sin(0))= sqrt(rg)

So, vb/vf= sqrt((sin(20 deg)+ cos(20 deg))/(cos(20 deg)-sin(20 deg)) = ~~ 1.464 Picture at http://s7.postimage.org/vx1nmm12z/image.png

Let N be normal reaction by ground on car let v be max velocity

along y-axis Ncos20 = μNsin20 + mg mg = N(cos20 -μsin20) .......(i)

along x-axis (mv^2)/r = Nsin20 + μNcos20 (mv^2)/r = N(sin20 + μcos20) .....(ii)

(ii)/(i)

(sin20 +μcos20)/(cos20 - μsin20) = (v^2)/(rg)

(tan20 + μ)/(1-μtan20) = v^2/rg

v^2 = rg(tan20 + μ)/(1-μtan20)

Let u be max velocity

μmg = mu^2/r u^2 = μrg

(v^2)/(u^2) = (tan20 + μ)/μ(1-μtan20)

on putting values we can get v/u

This is a simple banked curve with friction problem. The key derivation involved is the following formula:

$F_{net} = F_c$

$F_{net} = \frac {\tan \theta + \mu}{1 - \mu \tan \theta} mg = \frac {mv^2}{r}$

And we know that for a flat surface the equation is $v^2 = gr$ . The rest of the problem is trivial number chugging.

The derivation is left as an exercise (plus, it's a bit messy to write out). Hint: The frictional force will act in a downwards direction relative to the ramp. Thus, it will have a negative y-component and a positive x-component. Find the magnitude of the Normal force using the fact that $\sum F_y = 0$ .

First let us define all the forces acting on the bank and the car :

• $F_n$ = mg
• $F_fr$ = μ $F_n$
• $F_c$ = $\frac{mv^{2}}{r}$

These forces can be breakdown into x (horizontal) and y (vertical) components. In vertical direction we write :

• $F_n$ cosθ = mg + $F_fr$ sinθ
• $F_n$ cosθ - μ $F_n$ sinθ = mg
• $F_n$ = $\frac{mg}{cosθ - μsinθ}$

Now, in horizontal direction we write :

• $F_n$ sinθ + $F_fr$ cosθ
• $F_n$ sinθ + μ $F_n$ cosθ
• $F_n$ (sinθ + μcosθ)

Substitute the value of $F_n$ from the vertical direction:

• $\frac{mg}{cosθ - μsinθ}$ (sinθ + μcosθ)
• $\frac{(mg)(sinθ + μcosθ)}{cosθ - μsinθ}$

Since $F_n$ = $F_c$

• $\frac{(mg)(sinθ + μcosθ)}{cosθ - μsinθ}$ = $\frac{mv^{2}}{r}$

  Cancel out both mass (m) :

• $v^{2}$ = $\frac{(rg)(sinθ + μcosθ)}{cosθ - μsinθ}$

• v = $\sqrt{\frac{(rg)(sinθ + μcosθ)}{cosθ - μsinθ}}$

So by substituting all the values :

• $v_b$ = $\sqrt{\frac{(300 \times 9.8)(sin20 + (1)cos20)}{cos20 - (1)sin20}}$
• $v_b$ = 79.40
• $v_f$ = $\sqrt{\frac{(300 \times 9.8)(sin0 + (1)cos0)}{cos0 - (1)sin0}}$
• $v_b$ = 14 $\sqrt{15}$

So for $\frac{v_b}{v_f}$ : $\frac{v_b}{v_f}$ = $\frac{79.40}{14 \sqrt{15}}$

• $\frac{v_b}{v_f}$ = $\boxed{1.46}$