# Circular motion problem.

1. Sep 19, 2004

### 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 whats actually going on here, ive tried resolving in as many dimensions as i can but im just not getting anywhere. Think im starting to go insane.

2. Sep 19, 2004

### HallsofIvy

Staff Emeritus
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.

3. Sep 19, 2004

### Musiq

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

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

4. Sep 19, 2004

### saltrock

I think you need to use this formula,

Tan(theta)=V^2/rg
you have got theta ,that is 20 degrees v(you gotta 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,theres a topic called"help me out ASAP ,questions about circular moton' If you can answer my question i'll be really grateful.

5. Sep 19, 2004

### saltrock

Alternatively

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

6. Sep 19, 2004

### 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.

7. Sep 19, 2004

### HallsofIvy

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

8. Sep 20, 2004

### Musiq

Thanks for the help all.

Think ive 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 ya monkeys uncle.