Roller coaster in a circular loop problem

[razis_wwf13]

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.
F_c = 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 = F_c - mg = mv²/R - mg.

For c), i don't know how to attempt the qn. Hope you guys can help thx!

 

[Fightfish]

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.

 

[tomcue]

I would like to commend you for using the appropriate equations to solve this problem. Your solution for part (a) is correct, as the minimum speed needed for the car to maintain contact with the track at the top of the loop is 2√(gR).

For part (b), you are correct in assuming that the force vector needed is the normal force. At the top of the loop, the normal force is equal to the centripetal force, which is mv^2/R. Therefore, the force vector acting on the car by the track at the top of the loop is mv^2/R - mg.

For part (c), we can use the same equation as in part (a), but this time we are solving for the radius of the loop at which the car ceases to maintain contact with the track. Setting the speed to be 4v/5, we get 0.5mv1^2 + mg(2R) - 0.5mv2^2 - mg(0) = 0, where v1 is the speed at the bottom of the loop and v2 is the speed at the top. Rearranging this equation, we get R = (v1^2 - v2^2)/2g. Substituting v1 = 4v/5 and v2 = 2v, we get R = 3v^2/10g. Therefore, at a radius of 3v^2/10g, the car will cease to maintain contact with the track at the bottom of the loop.