How Do You Calculate Maximum Speed to Prevent a Car from Skidding on a Track?

  • Thread starter Thread starter jigs90
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on calculating the maximum speed of a car to prevent skidding on a circular track and over a hill. The car has a mass of 474 kg, a track radius of 197 m, and a coefficient of static friction of 0.123. The centripetal force equation, F = mv²/r, is essential for determining the maximum speed, where the normal force (N) equals the weight of the car (mg). The correct application of static friction is crucial for solving both scenarios, ensuring that the car maintains contact with the track.

PREREQUISITES
  • Centripetal force concepts
  • Static friction and its role in motion
  • Newton's second law of motion
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of the centripetal force equation, F = mv²/r
  • Learn about the role of static friction in circular motion
  • Explore the effects of different coefficients of friction on vehicle dynamics
  • Investigate the physics of motion on inclined planes and hills
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, and anyone interested in understanding vehicle dynamics and safety on curved tracks.

jigs90
Messages
19
Reaction score
0
Centrepetal Problem help PLEASE!

Homework Statement


I don't even know where to start on either of these two problems

A car of mass 474 kg travels around a flat,
circular race track of radius 197 m. The co-
efficient of static friction between the wheels
and the track is 0.123.
The acceleration of gravity is 9:8 m=s2 :
What is the maximum speed v that the car
can go without flying off the track?

It has a follow up too which I would really appreciate some help on!

The same car now travels on a straight track
and goes over a hill with radius 101 m at the
top.
What is the maximum speed that the car
can go over the hill without leaving the road?


A step by step outline would make my day!
Thanks


Homework Equations



A= V^2/ r
F= MV^2/ r


The Attempt at a Solution



I'm not sure how to go about tackling that problem, I tried just substituting 101 m in for the radius but that doesn't work.
 
Physics news on Phys.org
I tried doing the f= mv^2/r and when I entered in my answer for the first problem, it was wrong, I don't know how to take into consideration the static friction because I then used the equation mus (static friction) times N and subtracted that from the weight and then reworked the problem but my answer was still incorrect
 
jigs90 said:
A car of mass 474 kg travels around a flat,
circular race track of radius 197 m. The co-
efficient of static friction between the wheels
and the track is 0.123.
The acceleration of gravity is 9:8 m=s2 :
What is the maximum speed v that the car
can go without flying off the track?

Both problems will involve the centripetal force equation, but centripetal force is not a physical force, so in each situation, you need to think about what force provides the centripetal force.

For this first one, the car is on a horizontal curve. What force is acting on the tire surfaces to hold it on the turn? How do you calculate that force? That is the force which provides the centripetal force, so you can set it equal to F_centripetal.


The same car now travels on a straight track
and goes over a hill with radius 101 m at the
top.
What is the maximum speed that the car
can go over the hill without leaving the road?

In this problem, the centripetal force is toward the center of the hill, which we are to take as having a semicircular cross section. At the top of the hill, what forces are acting on the car? The centripetal force will be equal to the net force on the car. What is true about these forces when the car is on the verge of losing contact with the road surface when it is moving fast enough?
 
on the first one I set the equation of
mus(static)N = mv^2/ r but that still isn't right...Isn't the static friction providing the centripetal force?
 
jigs90 said:
on the first one I set the equation of
mus(static)N = mv^2/ r but that still isn't right...Isn't the static friction providing the centripetal force?

It does sound weird to use static friction when talking about a moving car, but consider that, in order to hold the turn, we do not want the tires to slide along the road surface. (Likewise, when we deal with rolling wheels, we will also use static friction rather than kinetic friction.)

So, yes, the static friction is providing the centripetal force for a level road curve. How do you find N?
 
isn't N equal to mg?
 
jigs90 said:
isn't N equal to mg?

Yes, for an object on a horizontal surface. What do you get when you put these pieces together?
 
isn't that what I had been using before though?
mus(static) mg= mv^2/s and the m's cancel and you solve for v^2 but I still keep getting it incorrect
 
jigs90 said:
isn't that what I had been using before though?
mus(static) mg= mv^2/s and the m's cancel and you solve for v^2 but I still keep getting it incorrect

What are the values you're putting into calculate v^2?
 
  • #10
.123(474)9.8
 
  • #11
jigs90 said:
.123(474)9.8

What is the radius of the curve?
 
  • #12
Nevermind...I figured it out. I was entering in the wrong decimal place. Thanks for all your help though. :)
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
15
Views
4K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 7 ·
Replies
7
Views
6K
Replies
13
Views
5K