Motion and force along a curved path-angle of curves?

Click For Summary
SUMMARY

The discussion focuses on calculating the banking angle required for a 740 kg car to navigate 175 m radius curves at a speed of 85 km/h without relying on friction. The key equations used include centripetal acceleration (a = v^2/r) and the relationship between tension and forces (Tcos(theta) = mv^2/r). The conclusion emphasizes that the frictional force must equal zero, necessitating a specific banking angle to ensure the normal force acts in the correct direction.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula (a = v^2/r)
  • Knowledge of free body diagrams and force resolution
  • Familiarity with the concept of normal force and its components
  • Basic grasp of Newton's second law (F = ma)
NEXT STEPS
  • Calculate the banking angle using the formula derived from Tcos(theta) = mv^2/r
  • Explore the implications of frictionless banking on vehicle dynamics
  • Investigate the effects of varying speeds on the required banking angle
  • Study real-world applications of banking angles in racetrack design
USEFUL FOR

Physics students, automotive engineers, and anyone interested in the dynamics of vehicles on curved paths will benefit from this discussion.

a18c18
Messages
21
Reaction score
0

Homework Statement



An automobile club plans to race a 740 kg car at the local racetrack. The car needs to be able to travel around several 175 m radius curves at 85 km/h. What should the banking angle of the curves be so that the force of the pavement on the tires of the car is in the normal direction? (Hint: What does this requirement tell you about the frictional force?)

Homework Equations



a=v^2/r
Tcos(theta)=mv^2/r
F=ma

The Attempt at a Solution



Friction must equal 0?
 
Physics news on Phys.org
Correct. So draw a free body diagram for the situation.

Here is a hint : Resolve the normal force into x and y components. One of these components should equal the weight of the car, the other should provide the required centripetal force towards the center of the track.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
4
Views
5K
Replies
9
Views
4K
Replies
16
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
Replies
5
Views
2K
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
10K
  • · Replies 1 ·
Replies
1
Views
3K