How Does Friction Affect a Car's Acceleration on a Curved Track?

[fizz123]

A car at the Indianapolis-500 accelerates uniformly from the pit area, going from rest to 340 km/h in a semicircular arc with a radius of 196 m.

Determine the tangential acceleration of the car when it is halfway through the turn, assuming constant tangential acceleration.
7.2429 m/s2
Determine the radial acceleration of the car at this time.
22.75 m/s2
If the curve were flat, what would the coefficient of static friction have to be between the tires and the roadbed to provide this acceleration with no slipping or skidding?


I've solved the first two parts, but I'm stuck on the third.
I tried Force of Static Friction = Coefficient * Force of Normal and then setting that equal to the Centripetal Force ((m * V^2) / R), ending up with
Coefficient = V^2 / (R * g)

Don't know what else to do...

 

[maverick280857]

Flat? Do you mean there was a banking angle involved?

 

[OlderDan]

I tried Force of Static Friction = Coefficient * Force of Normal and then setting that equal to the Centripetal Force ((m * V^2) / R), ending up with
Coefficient = V^2 / (R * g)

Don't know what else to do...

You know R. You know the tangential acceleration is constant. Figure out what v is from the tangential acceleration and the distance travelled.

 

[fizz123]

what should i do after i solve for v? setting the force of friction equal to the centripetal force doesn't work

and, flat means no banking

 

[OlderDan]

There may be a bit more to this than first meets the eye. I was just looking at your derived equation for the coefficient when I said to find the velocity, but you must have found the velocity to get the radial acceleration in the first place. I'm now thinking your result for the coefficient is coming out a bit too small, and I think I know why. What are you getting? What do you think is right?

Try finding the resultant of the tangential acceleration and the radial acceleration and using that to compute the coefficient of friction.

 

[fizz123]

i've tried .5807 and .7383, but i don't remember how i got those values. After getting the resultant acceleration, do i use the equation: sum of forces = m*a ? and would the forces be force of friction and the centripetal force?

 

[OlderDan]

The centripetal force comes from the frictional force. So does the force that provides the tangential acceleration. The only force being applied to the car to give it an acceleration is friction. If your calculated accelerations are correct (I think they are) you have a resultant acceleration (from summing the two as vectors) that is far in excess of g. It is somewhat more than the radial acceleration alone. The only way you can get that is if the coefficient of friction is much greater than 1. Apparently those tires are really sticky. In the horizontal direction, "sum of forces" is the force of friction and it is equal to m*a.

 

[fizz123]

thank you, it ended up being 2.44