How High Can a Roller Coaster Start Without Flying Off the Track?

[dekoi]

A roller coaster car on the frictionless track shown, starts from rest at height h. The track is straight until point A. Between points A and D, the track consists of circle-shaped segments of radius R.

Question:
What is the maximum height hmax from which the car can start so as not to fly off the track when going over the hill at point C? Give you answer in terms of the radius R.
Hint: This is a two-part problem. First find vmax at C.

My answer:
I have tried many different methods, but cannot find the right one. The answer in the textbook tells me that hmax = 3/2R.
I know how to find hmax once I have vmax, but the problem is finding vmax in the first place.
I don't know what I'm supposed to do with the 30 degree angle. I thought of using it as the angular position, but I can't seem to continue from there to find vmax. I calculated that the angular position would be 60 degrees (90 - 30).
Does the fact that there are 2 circles have something to do with my answer? Or is that just there to confuse me?
I also tried drawing a FBD at point C. The only two forces I obtained were a normal force (up) and gravity (down). However, because it is a circle, is my normal force supposed to be pointing down also (toward the centre of the circle)? Does this have something to do with my answer also?
Please help-- I have tried many different methods but cannot seem to get the right answer.

 

[Doc Al]

Consider that if the car is to maintain contact with the track at point C, what kind of motion must it execute? (Hint: Is it accelerating?) How do you specify the condition for "just about to lose contact"? (Hint: What happens to the normal force?) Apply Newton's 2nd law to the car at point C. (What's the net force on the car?)

 

[dekoi]

I do not know what the net force is on the car. That's what I was trying to figure out with the free-body diagram. I'm assuming the net force is toward the centre of the circle since the car is undergoing circular motion at point C.

If the car is to maintain contact with the track at point C, it must execute circular motion which means it's accelerating toward the centre of the circle. Right?

I don't know what happens to the normal force when the car is just about to lose contact. Can you help?

 

[Doc Al]

I do not know what the net force is on the car. That's what I was trying to figure out with the free-body diagram. I'm assuming the net force is toward the centre of the circle since the car is undergoing circular motion at point C.

Right.

If the car is to maintain contact with the track at point C, it must execute circular motion which means it's accelerating toward the centre of the circle. Right?

Exactly! This is key: use what you know about centripetal acceleration.

I don't know what happens to the normal force when the car is just about to lose contact.

Here's a hint: What happens to the normal force when there is no contact?

 

[dekoi]

When there is no contact, the normal force is 0, so only gravity is taken into account. I'm assuming this is what I have to use in order to find the acceleration? And just for some clarification before I attempt this problem again: Does the angle have anything to do with my answer? Or even that fact that there are two circles...

 

[Doc Al]

When there is no contact, the normal force is 0, so only gravity is taken into account.

Right. That's what you need to find the net force.

And just for some clarification before I attempt this problem again: Does the angle have anything to do with my answer? Or even that fact that there are two circles...

All that matters is that point C is on the top of a circle. Use that fact to determine the acceleration at point C.

 

[dekoi]

Thank you.

I will attempt the problem and post again if I don't get the correct answer.