Roller Coaster Question - Please Help

[ntukza]

A roller coaster freewheels down a frictionless track from rest a height h above the ground. At the bottom the track becomes a vertical loop with diameter 2a = 16m. How high above the ground (value of h) must the roller coaster start if it is to reach the top of the vertical loop without lifting off the track?

 

The given answer is 20m. I would have thought that the roller coaster needs to start at a height equal to, or slightly more than, the diameter of the vertical circle at the bottom. Energy at start = energy at end, therefore mgh = mg2a, and h = 2a = 16m. Please help me understand why it is not so.

 

Thank you in advance

[swansont]

What would be the speed at the top of the loop if you started at 16m? Would the car stay on the track under that condition?

[hermanntrude]

roller coasters never lift off the track because they're designed not to. I hate unrealistic questions

[swansont]

roller coasters never lift off the track because they're designed not to. I hate unrealistic questions

 

It explicitly states that it's freewheeling. This "unrealistic" part actually shows why that design is necessary.

[A Childs Mind]

roller coasters never lift off the track because they're designed not to. I hate unrealistic questions

 

 

ummm not to be rude i mean this out of respect. i do belive you miss understand the question. hes stating. that how high must the rolercoster be of the ground without being moved of the tracks. cuz noing me. id say somethng smart and say just pickit up and move it to the point it needs to go

[ntukza]

The way I see it, if the roller coaster starts at 16m above the ground, its mechanical energy is mgh and obviously v=0. It will therefore reach the top of the vertical loop of diameter 16m but will not go past the top point because there v=0 and E=mgh again, so it will fall off. For it to remain on the tracks h must be greater than 16. This means even h=17m will be sufficient. Why does h have to be 20m. Can someone please explain to me

[Kyrisch]

I'm pretty sure you have to factor in centripetal force when the normal force is zero (i.e. when the cars are just barely touching the tracks). [math] F_c = m\frac{v^2}{ R }[/math]

[swansont]

Exactly — there's a condition that the car is moving is a circle, so the force toward the center must be mv^2/r, and that can't hold if v=0.

[ntukza]

Ok, it's starting to make sense now. So the centripetal force needs to act away from the centre of the circle so as to balance the weight which acts towards the centre of the circle?

 

mg = m(v^2)/r

 

v^2 = g.r

 

So we can find v at the top, and then what do we do from there?

[swansont]

Ok, it's starting to make sense now. So the centripetal force needs to act away from the centre of the circle so as to balance the weight which acts towards the centre of the circle?

 

mg = m(v^2)/r

 

v^2 = g.r

 

So we can find v at the top, and then what do we do from there?

 

The centripetal force acts toward the center of the circle, which is why you can equate it with the gravitational force.

 

So now you know what speed it has at the top. What height must you start at for it to have that speed?

[Kyrisch]

Ok, it's starting to make sense now. So the centripetal force needs to act away from the centre of the circle so as to balance the weight which acts towards the centre of the circle?

 

mg = m(v^2)/r

 

v^2 = g.r

 

So we can find v at the top, and then what do we do from there?

 

Actually, you are mistaked here. The centripetal force is actually towards the center of the loop. The force of gravity and the centripetal force are acting the same direction. The only force opposing is the normal force. Thus, [math]F_N = F_C + F_g[/math], but with the constraint that the normal force is zero, it simplifies to what you have written up there, that the two are equal (but opposite).

 

Further, because you don't have any figures for velocity at all, you should try to use the equations you have to substitute velocity out of the equation you wish to solve. The point is just that you now have another constraint in your system.

[ntukza]

Actually, you are mistaked here... The force of gravity and the centripetal force are acting the same direction... the two are equal (but opposite).

 

If they are opposite then surely they can't both point in the same direction, right?

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I've figured it out guys, thanks for your help. The only thing I don't fully understand is why weight = centripetal force if centripetal force always acts toward the centre of the circle. But assuming it does:

 

mg=m(v^2)/r

 

v^2 = g.r = 80

 

Now the energy equations:

 

mgh = mg(2a) + 1/2m(v^2)

 

for 2a=16 and (v^2)=80, h=20

 

Once again thanks for helping me see that I needed to consider centripetal force

[swansont]

If they are opposite then surely they can't both point in the same direction, right?

 

Re-read why Kyrisch wrote — you had set them to be opposite, and Kyrisch was correcting that.

 

 

I've figured it out guys, thanks for your help. The only thing I don't fully understand is why weight = centripetal force if centripetal force always acts toward the centre of the circle.

 

graqvity is the centripetal force ONLY at the top of the loop, and in the condition that there is no normal force. In general, gravity will be one contributor to the centripetal force at the top and bottom of the loop, because those are the only places it is directed along the radius. At other places it will contribute both a radial and a tangential acceleration, so the object will be changing speed. (Conservation of energy also tells you this, so you can use either one to see how the speed changes.)