What formula is most suitable?

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In summary, @Nugatory doesn't know which formula would suit best for the gondola (pictured), but he believes that it would be huge due to the rotational inertia. He is unsure of what axis the gondola rotates about, but he thinks that it's the arm/pendulum-tube that it's attached to. He needs to know how high the axis is, and then he can use the "hoop about a diameter" formula to find the mass and radius of the gondola. He also notes that an external torque provider, such as a motor, would be used to cause the gondola to spin.
  • #1
solarmidnightrose
28
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Homework Statement
I've been trying to calculate the rotational inertia of this rotating gondola on this ride at this theme park.

However, they do not directly give me a specific formula to use-because with different objects, there are different formulae that you use to determine the rotational inertia of an object

mass of gondola = 6 tonne (6000kg)
radius = 3m


I will attach a picture of what it looks like
Relevant Equations
I have attached a picture of different possible formulae's
Picture of gondola:
244604


Formulae:
244603


Initially, I used the equation: I=mr^2
I=(6000)x(3)^2​
T=54,000kgm^2
(but apparently this is incorrect)
So, basically, I don't know which formula would suit best for the gondola (the picture above). I would consider the gondola a hollow object as the mass is concentrated on the outer edges.
 
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  • #2
solarmidnightrose said:
which formula would suit best for the gondola
The key question is what axis is it rotating about? I am unfamiliar with this equipment, but it looks like it swings like a pendulum from somewhere up above. If so, you need to know how high that is. You would then apply the "hoop about a diameter" formula, and using the parallel axis theorem add a term mh2 where h is the height of the axis.
 
  • #3
The assumption that all the weight is in a ring is massive.
It will affect the torque required to drive the system, and the velocity of the driven system.
Furthermore, If all the mass were in the ring it would have to be very precisely balanced.
 
  • #4
haruspex said:
The key question is what axis is it rotating about? I am unfamiliar with this equipment, but it looks like it swings like a pendulum from somewhere up above.
Yes, you're assumption/idea of the ride is correct :) (sorry, I should have provided a video)
video: (I believe only the first 10 seconds is enough)

Anyways, if I take into consideration the axis that the gondola is rotating about then that would be the arm/pendulum-tube that the gondola is attached to? (I've circled this part in blue) Is this correct?-I just want some confirmation please.
244611

haruspex said:
If so, you need to know how high that is. You would then apply the "hoop about a diameter" formula, and using the parallel axis theorem add a term mh2 where h is the height of the axis.
The length of the pendulum tube is 15m.
Would it be alright to say that the 'height' of the axis (i.e. the pendulum tube) is 15m?

If I use the "hoop about a diameter formula", then I would:
  1. substitute the mass/radius of the gondola into the "hoop about a diamter formula"
  2. add the (mh^2) to the above answer.
My attempt at your suggestion:
244612

@haruspex , am I doing this correctly? have I substituted the correct values into the formula? have I even done the right thing by adding "mh^2" to "hoop about a diameter formula"?

Thanks for your help/input; very kind.
 
  • #5
tkyoung75 said:
The assumption that all the weight is in a ring is massive.
Thanks for that, @tkyoung75 .
My teacher was telling us that for this report that many assumptions would have to be made in order to help explain the physics concepts behind these rides at theme parks.

So, since this rotational inertia of this gondola must be HUGE, then is that why an external torque provider such as a motor would be used in rides such as this to get it spinning in the first place? If so, then that would really help my report.

Thanks for your help :)
 
  • #6
solarmidnightrose said:
So, basically, I don't know which formula would suit best for the gondola (the picture above).
The formulas you posted are all different special cases of the general and always applicable ##I=\int \rho(\vec{r})r^2dV## (and it would be a good exercise to see if you can derive some of them from this general formula).

You'll solve this problem by making a reasonable assumption about what ##\rho## might be, identifying the axis of rotation, and then evaluating the integral.
 
  • #7
Nugatory said:
The formulas you posted are all different special cases of the general and always applicable ##I=\int \rho(\vec{r})r^2dV## (and it would be a good exercise to see if you can derive some of them from this general formula).

You'll solve this problem by making a reasonable assumption about what ##\rho## might be, identifying the axis of rotation, and then evaluating the integral.
I'm sorry, but my teenage brain is screaming/crying looking at that integral.

Also, what is this 'roh' in the integral?

I apologise @Nugatory if I come off as rude, but this is something that I haven't been taught yet. Do you think it's possible to learn all this derivative-stuff/integration in a couple of hours (6hrs max)? If yes, then "yay!".

If no, then I'm screwed.
 
  • #8
What do you have to do to change the direction?

solarmidnightrose said:
So, since this rotational inertia of this gondola must be HUGE, then is that why an external torque provider such as a motor would be used in rides such as this to get it spinning in the first place? If so, then that would really help my report.

This about Newtons laws of motion.
 
  • #9
tkyoung75 said:
What do you have to do to change the direction?
Apply an unbalanced torque to cause it to accelerate angularly-i.e. spin/rotate. (Newtons II Law).

tkyoung75 said:
This about Newtons laws of motion.
Oh okay, that's cool.

Thanks-I believe I get this idea of rotational inertia better now.

It's just that I have a hard time trying to actually calculate the rotational inertia (I guess you could say I'm not very mathematically strong).
 
  • #10
Do you know the basic formula for inertia?
 
  • #11
tkyoung75 said:
Do you know the basic formula for inertia?
Is it I=mr^2? (rotational inertia)
 
  • #12
That is the one I had in mind. Do you know how to add them together, if you have a mass in the centre and a mass on the outside?
 
  • #13
tkyoung75 said:
That is the one I had in mind. Do you know how to add them together, if you have a mass in the centre and a mass on the outside?
No, I have no idea.

But I have a feeling its got something to do with Centre of Mass? (which is something I can do-calculate the CoM; but I don't know if that's relevant).

An earlier post mentioned something about the Parallel Axis Theorem-(but I have never come across this before) and I am not sure how important I need to know this in order to find the rotational inertia of the gondola in this ride.
 
  • #14
solarmidnightrose said:
Yes, you're assumption/idea of the ride is correct :) (sorry, I should have provided a video)
video: (I believe only the first 10 seconds is enough)

Anyways, if I take into consideration the axis that the gondola is rotating about then that would be the arm/pendulum-tube that the gondola is attached to? (I've circled this part in blue) Is this correct?-I just want some confirmation please.
View attachment 244611

The length of the pendulum tube is 15m.
Would it be alright to say that the 'height' of the axis (i.e. the pendulum tube) is 15m?

If I use the "hoop about a diameter formula", then I would:
  1. substitute the mass/radius of the gondola into the "hoop about a diamter formula"
  2. add the (mh^2) to the above answer.
My attempt at your suggestion:
View attachment 244612
@haruspex , am I doing this correctly? have I substituted the correct values into the formula? have I even done the right thing by adding "mh^2" to "hoop about a diameter formula"?

Thanks for your help/input; very kind.

Yes, that looks about right... except, watching the video to the end I see there is a long section of tube above the axle, possibly with a counterweight on the end. This could easily add 50%; hard to know without more data.

Also I see there is some rotation about the tube axis, but this is small by comparison and does not add to the above since it relates to an orthogonal rotation.
 
  • #15
solarmidnightrose said:
Parallel Axis Theorem-(but I have never come across this before)
So Google it.
 
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  • #16
Well if you had two masses each traveling in the same direction and they hit a wall at the same speed at the same with force F, what would the total force be?
 
  • #17
haruspex said:
Yes, that looks about right... except, watching the video to the end I see there is a long section of tube above the axle, possibly with a counterweight on the end. This could easily add 50%; hard to know without more data.

Also I see there is some rotation about the tube axis, but this is small by comparison and does not add to the above since it relates to an orthogonal rotation.
I am so thankful for your help
haruspex said:
So Google it.
haha, yeah I did.
 

Related to What formula is most suitable?

1. What factors should be considered when determining the most suitable formula?

There are several factors that should be considered when determining the most suitable formula for a particular experiment or problem. These include the purpose of the formula, the variables involved, the accuracy and precision required, and any constraints or limitations.

2. How do I know if a formula is appropriate for my experiment?

A formula can be considered appropriate for an experiment if it accurately represents the relationship between the variables being studied and if it produces results that align with the goals and objectives of the experiment. It should also be applicable to the specific conditions and parameters of the experiment.

3. What is the difference between a theoretical formula and an empirical formula?

A theoretical formula is derived from mathematical principles and theoretical models, while an empirical formula is based on observations and data collected from experiments. Theoretical formulas are often more general and can be used to make predictions, while empirical formulas are more specific and may only be applicable to a certain set of conditions.

4. Are there any guidelines for choosing a formula?

There are no set guidelines for choosing a formula, as it ultimately depends on the specific experiment and the goals of the scientist. However, it is important to thoroughly understand the variables and relationships involved, and to consider the accuracy, precision, and limitations of the formula before selecting it for use.

5. Is it possible to modify existing formulas for better results?

Yes, it is possible to modify existing formulas to improve their accuracy or applicability to a specific experiment. This may involve adjusting constants or coefficients, incorporating additional variables, or combining multiple formulas. However, any modifications should be carefully tested and validated before using in a scientific study.

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