Circular motion and g forces in rollercoaster

[Erucibon]

Homework Statement

A trolley goes down a slope and then into a vertical circle, and there must be less than 5g at the top and bottom of the loop. Friction is ignored. Determine a suitable height, radius that satisfies these conditions.

Relevant Equations

Ki + Ui = Kf + Uf
Gforce = (a+g)/g
a = v^2/r

I my attempt, I set the drop height to 20m and using conservation of energy, i calculated the speed at the bottom. Calculating centripetal acceleration, if the radius of the circle is less than 10m then the g force is greater than 5, if equal to 10m the velocity at the top is 0 and there is 0 acceleration, and if radius is more than 10m then the velocity squared at the top is negative. When calculating the velocity at the top of the circle, i used conservation of energy. Could someone please point out where I went wrong or if this approach is completely wrong

 

[haruspex]

if radius is more than 10m then the velocity squared at the top is negative

That is not the only consideration. Consider the forces acting on it at the top. You don't want it to fall off the track.

Also, you seem to have approached it backwards, plugging in numbers at the start. Create unknowns and work entirely algebraically to find the relationship required between drop height and radius.

 

[Erucibon]

That is not the only consideration. Consider the forces acting on it at the top. You don't want it to fall off the track.

Also, you seem to have approached it backwards, plugging in numbers at the start. Create unknowns and work entirely algebraically to find the relationship required between drop height and radius.

I worked out a relationship here. I am not sure what you mean by the forces at the top and i think the top is where i messed up. Sorry I forgot to attach my working.

 

[haruspex]

I worked out a relationship here. I am not sure what you mean by the forces at the top and i think the top is where i messed up. Sorry I forgot to attach my working.

What forces act on it at the top? What is its acceleration at the top if it is staying on the track? What equation does that give you?

 

[Cutter Ketch]

I worked out a relationship here. I am not sure what you mean by the forces at the top and i think the top is where i messed up. Sorry I forgot to attach my working.

Using conservation of energy you correctly worked out a relation between h and r: in order to satisfy the requirement that the max acceleration be < 4g, h must be < 2r. However, that does not determine h or r. The 4g requirement is satisfied for any pair of h and r that satisfy h < 2r. However instead of introducing your second constraint (that the car not fall off the loop). You first arbitrarily picked h and r. No! The second constraint won’t be satisfied for any choice of h and r. You must continue to treat h and r as variable parameters and keep doing algebra.

As haruspex has been suggesting, you have to formulate the physics description of that second requirement. What does it mean that the car doesn’t come off the road at the top of the loop? What are the forces? What force is required? What constraint does that put on v and r and h?

 

[Erucibon]

Using conservation of energy you correctly worked out a relation between h and r: in order to satisfy the requirement that the max acceleration be < 4g, h must be < 2r. However, that does not determine h or r. The 4g requirement is satisfied for any pair of h and r that satisfy h < 2r. However instead of introducing your second constraint (that the car not fall off the loop). You first arbitrarily picked h and r. No! The second constraint won’t be satisfied for any choice of h and r. You must continue to treat h and r as variable parameters and keep doing algebra.

As haruspex has been suggesting, you have to formulate the physics description of that second requirement. What does it mean that the car doesn’t come off the road at the top of the loop? What are the forces? What force is required? What constraint does that put on v and r and h?

What forces act on it at the top? What is its acceleration at the top if it is staying on the track? What equation does that give you?

At the top fnet=ma=Fc=N+mg=ma(centripetal)
To not come off, N>= 0
mac - mg >= 0
ac>=g
v^2/r >= g
v >= root(rg)
Using conservation of energy between top and bottom of loop
v(bottom) = root(5rg)
If this is correct, I'm not sure what to do next

 

[Lnewqban]

Using low enough height and a huge radius, you could easily satisfy the limiting condition of less than 5g at the bottom of the loop.
The problem with that combination is that the car would not have enough kinetic energy to complete the loop, falling down from the top.
For that reason, the relation $h<2r$ shown in your calculations would not work, since the car would have zero velocity around the top of the loop.

You need to determine values of height and radius that provide the car with enough energy to complete the loop, but not too much velocity (exceeding 5g total acceleration) at the bottom of the loop.
Determining the minimum velocity possible at the top of the loop is the beginning of the solution.

 

[Erucibon]

Using low enough height and a huge radius, you could easily satisfy the limiting condition of less than 5g at the bottom of the loop.
The problem with that combination is that the car would not have enough kinetic energy to complete the loop, falling down from the top.
For that reason, the relation $h<2r$ shown in your calculations would not work, since the car would have zero velocity around the top of the loop.

You need to determine values of height and radius that provide the car with enough energy to complete the loop, but not too much velocity (exceeding 5g total acceleration) at the bottom of the loop.
Determining the minimum velocity possible at the top of the loop is the beginning of the solution.

When i calculate the g force for this, is it equal to (a+g)/g at the bottom and (a-g)/g at the top, where a is the centripetal acceleration? Or is it just a/g (ac = a net ?)

 

[haruspex]

At the top fnet=ma=Fc=N+mg=ma(centripetal)
To not come off, N>= 0
mac - mg >= 0
ac>=g
v^2/r >= g
v >= root(rg)
Using conservation of energy between top and bottom of loop
v(bottom) = root(5rg)
If this is correct, I'm not sure what to do next

Much better.
So what is the (minimum) normal force at the bottom?

 

[haruspex]

Determining the minimum velocity possible at the top of the loop is the beginning of the solution.

Hasn't that been done in post #6?

 

[Cutter Ketch]

v(bottom) = root(5rg)

Yes, you got it. I believe that’s right. (checking in my head can be unreliable!) Now, you have already shown you understand how v bottom is related to h, right? Substitute here and you have a relation between h and r from the second constraint. You already had a relation between h and r from the first constraint. Two equations in two unknowns. The rest is algebra. What’s the maximum h and r that satisfy both inequalities?

 

[Cutter Ketch]

Hey! Wait a minute! In your calculations you required that the force be less than 4 g’s. The problem said 5 g’s. Your first constraint is a little wrong.

 

[Lnewqban]

When i calculate the g force for this, is it equal to (a+g)/g at the bottom and (a-g)/g at the top, where a is the centripetal acceleration? Or is it just a/g (ac = a net ?)

For bottom of loop:
$Max~gforce=5=(a_c+g)/g$
$5g=a_c+g$
$a_c=5g-g$
$a_c=4g$

Now, if, as you have properly calculated

$V_{top}\geq\sqrt{gr}$

1) The car has both potential and kinetic energy when at top of the loop.
2) You can use the energy balance equation to calculate $h$, where the car has only potential energy and $V_{initial}=0$
3) Find the relation between $h$ and $r$
4) With the calculated value of $V_{bottom}$, where there is no potential energy (relative to initial height), you can verify whether or not the value of $a_{c~bottom}$ exceeds the limit value of $4g$

 

[haruspex]

Hey! Wait a minute! In your calculations you required that the force be less than 4 g’s. The problem said 5 g’s. Your first constraint is a little wrong.

The "g force" must be less than 5, but I believe that refers to the net of forces applied to the exterior, so excludes gravity itself. I'm not sure which calculation you are referring to in regard to "4 g's", but perhaps that referred to the net force including gravity.

Fwiw, it looks to me that the question setter has blundered and there is no solution.

 

[Cutter Ketch]

The "g force" must be less than 5, but I believe that refers to the net of forces applied to the exterior, so excludes gravity itself. I'm not sure which calculation you are referring to in regard to "4 g's", but perhaps that referred to the net force including gravity.

Fwiw, it looks to me that the question setter has blundered and there is no solution.

Oh, jeez. You’re right. Well, this may be a good time to “listen to what they meant rather than what they said“. If you take the maximum allowed force to be 5 “more” g‘s (so really 6 g’s total) you can at least get a solution.

You know, I once had a physics professor who purposely searched for all the problems in the textbook that were wrong one way or another. He figured we learned more that way.

 

[Cutter Ketch]

G-force is defined as the magnitude of forces excluding gravity, so, for example, it takes 1 g to stand still, and free fall is zero g’s, even though clearly you are still being acted on by gravity. Basically G-force isn’t the net force. It’s how much the forces are compressing you.

However, looking at the wording of the problem, it says “there must be less than 5g at the top and bottom of the loop”. The title of the post says “g forces” but it doesn’t actually mention “g forces” or “g’s” in the problem statement. Sure, it’s really a mistake, but it isn’t entirely ludicrous to interpret it to mean that the net acceleration should never exceed 5 g. I’m sure that’s what they had in mind, anyway.