Roller coaster in a circular loop problem

Click For Summary
SUMMARY

The roller coaster problem involves calculating the minimum speed required for a car to maintain contact with a circular loop of radius R. The minimum speed at the bottom of the loop is determined to be v = 2√(gR). When the car travels at a speed of 6v/5 at the bottom, the normal force at the top of the loop can be calculated using the equation N = mv²/R - mg. For a speed of 4v/5, the point where the car loses contact with the track can be found by analyzing the centripetal force as a function of position along the track.

PREREQUISITES
  • Understanding of centripetal force and its application in circular motion
  • Familiarity with Newton's laws of motion
  • Knowledge of energy conservation principles in physics
  • Ability to manipulate equations involving forces and motion
NEXT STEPS
  • Study the derivation of centripetal force in circular motion
  • Learn about energy conservation in mechanical systems
  • Explore the effects of varying speeds on forces in circular tracks
  • Investigate the conditions for maintaining contact in vertical loops
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of roller coasters and circular motion mechanics.

razis_wwf13
Messages
1
Reaction score
0

Homework Statement


A roller coaster car is on a track that forms a circular loop of radius R in the vertical plane.

(a) If the car is to maintain contact with the track at the top of the loop, what is the minimum speed v that the car must have at the bottom of the loop? Ignore air resistance and rolling friction.
(b) If the car has a speed 6v/5 at the bottom of the loop, then what will be force vector acting on the car by the track when it reaches the top of the loop? [v is the minimum speed calculated in (a).]
(c) If the car has a speed 4v/5 at the bottom of the loop, locate the point on the track where the car ceases to maintain contact with the track.


Homework Equations


0.5mv12 + mg(2R) - 0.5mv22 - mg(0) = 0.
Fc = N - mg


The Attempt at a Solution


For a), i just used the first eqn, sub in all the known datas, i got v = 2sqrt(gR) as the min speed needed.

For b), i assumed force vector needed is normal force. So at the top, i used the 2nd eqn, and i got N = Fc - mg = mv2/R - mg.

For c), i don't know how to attempt the qn. Hope you guys can help thx!
 
Physics news on Phys.org
Your answer for (a) looks dubious; recheck your working.
For (c), consider how the centripetal force on the car varies with its position on the track (hint: use an angle approach). Then find the point where N = 0.
 

Similar threads

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