What is the minimum speed required for a car to safely navigate a banked curve?

[Musiq]

having trouble with this problem, any help greatly appreciated.

A car moves in a horiztonal circle of radius 75m around a bend which is banked at an angle of 20 degrees to the horiztonal. At what speed should the car be driven if it is to have no tendency to slip?

Right so after just gettin my head around what's actually going on here, I've tried resolving in as many dimensions as i can but I am just not getting anywhere. Think I am starting to go insane.

 

[HallsofIvy]

There are two vectors here: the gravitational force on the car which is straight down and the centrifugal force (which will depend on the car's speed) which is directly inward toward the center of the circle. Find the sum of those two forces. The car will "have no tendency to slip" if the force is pependicular to the bank: in other words, is 20 degrees to the vertical.

 

[Musiq]

There are two vectors here: the gravitational force on the car which is straight down and the centrifugal force (which will depend on the car's speed) which is directly inward toward the center of the circle. Find the sum of those two forces. The car will "have no tendency to slip" if the force is pependicular to the bank: in other words, is 20 degrees to the vertical.

Would that be parallel to the horiztonal or parallel to the bank?

Thanks for the help. Although I am not sure how I am going to derive an actual figure for the speed still

 

[saltrock]

I think you need to use this formula,

Tan(theta)=V^2/rg
you have got theta ,that is 20 degrees v(you got to find it),you have got r and g=9.81

I think i have solved your problem.If you go back and have a look at the list of questions,there's a topic called"help me out ASAP ,questions about circular moton' If you can answer my question i'll be really grateful.

 

[saltrock]

Alternatively

V=rw
you have got r yo have got theta,change that theta into radians /sec,then you will get v

 

[saltrock]

This is what i think is right,make sure you check it to some experts answer before writting that down on examination or something like that.

 

[HallsofIvy]

The circle is horizontal. The vector toward the center of the circle is horizontal.

 

[Musiq]

Thanks for the help all.

Think I've got it now, resolved the centripetal force to be

9.81m tan 20

therefore, because f=(mv^2) / r

9.81m tan 20 = (mv^2)/r

masses cancel, bobs you monkeys uncle.