Maximum Velocity for a Banked Curve Without Sliding or Rolling Over

[sricho]

Determine the maximum velocity at which a car can travel if the banking angle is 15º for the following conditions
1) It is not to slide radially outwards.

2) It is not to roll over.

given:
coefficient of friction as 0.4
r = 60m
width of car is 2m and height from bank to centre of car is 0.5m

Homework Equations

Looking at some work i think the equation that i need to use is
V = square root (( tan feta + coefficent of friction/ 1-coefficent of friction tan feta) x r x g)

The Attempt at a Solution

my attempt at this solution gave me an answer of 22.9m/s so 23m/s
i wanted to check if this was correct and I am confused on why i have been given the length of the car and its distance from bank to centre of car.

 

[tiny-tim]

Welcome to PF!

Hi sricho! Welcome to PF!

Determine the maximum velocity at which a car can travel if the banking angle is 15º for the following conditions
…
2) It is not to roll over.

given:
coefficient of friction as 0.4
r = 60m
length of car is 2m and height from bank to centre of car is 0.5m
…
im confused on why i have been given the length of the car and its distance from bank to centre of car.

because it will roll over if the net torque about the inside wheel is towards the centre …

for that, you need to know where the centre of mass is

hmm … are you sure 2m isn't the width of the car?

 

[sricho]

yes sorry that is the width of the car. so i need to work out the centre of mass. rite that makes sense with the given lenghs.
was the equation for the velocity correct?
pretty new to this so any help would be ausome
thanks

 

[tiny-tim]

Hi sricho!

Sorry, I missed your reply.

… was the equation for the velocity correct?

Looking at some work i think the equation that i need to use is
V = square root (( tan feta + coefficent of friction/ 1-coefficent of friction tan feta) x r x g)

(have a theta: θ and a mu: µ )

Difficult to tell, since you haven't shown how you got it …

and do remember, the normal force won't be the usual mgcosθ, because of the centripetal acceleration

 

[sricho]

well i got my answer by using the equation for velocity and subsituting the values i have.
so
sqr(tan15 degree + 0.4 / 1 - 0.4 tan 15 degrees) x 60 x 9.8

but this doesn't use the normall reaction force and i don't think this answers the question

 

[sricho]

Ncos θ - µNsin θ = mg
N(cos θ - µsinθ) = mg
N = mg / cos θ - µsinθ

This is the equation i have for the normall reactiong force but you need the mass of the object where its not given.
so I am stuck?

 

[tiny-tim]

No, you don't need to know the mass: it always cancels.

And your equations haven't taken the centripetal acceleration into account.

 

[sricho]

sorry had a busy weekend
so how do i use that acceleration in the equation i have.
and what is the equation for the acceleration?
thanks
sam

 

[sricho]

is this the equation i need ?
Ac = - w^2 r
Ac = - v^2 / r

where w is omega

how do i add this into the velocity equation?

 

[tiny-tim]

… what is the equation for the acceleration?

you should be able to look that up

centripetal acceleration is v²/r (= ω²r)​

 

[sricho]

well I am just all confused now, is there any chance you could show me how its done , i would be very happy if you could. it will help loads with the other questions i have to do.