Maximum angular velocity from a swing carousel

Click For Summary

Homework Help Overview

The problem involves determining the maximum angular velocity of a swing carousel before the chain supporting the passenger breaks. Key parameters include the maximum tension the chain can withstand, the radius from the center of the ride, the maximum mass of the passenger, and the length of the chain.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between tension, gravitational force, and angular velocity, with some attempting to derive formulas based on equilibrium conditions. Others express confusion regarding the role of angles and forces in the system, and question the validity of certain assumptions.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem, including setting up equations of motion and considering energy conservation. Some have offered guidance on drawing free body diagrams to clarify the forces involved, while others are questioning the correctness of derived formulas and assumptions.

Contextual Notes

There are indications of missing information regarding the angle of the chain and its impact on the forces involved. Participants are also navigating the complexities of the problem, including the relationship between angular velocity and the changing radius as the carousel spins.

piisexactly3
Messages
36
Reaction score
0
The problem is to find out the maximum angular velocity that the swing carousel can spin at before the chain holding the passenger breaks. The maximum tension that the chain can withstand is known (T). The radius from the centre of the ride to the chain when stationary is known (r). The maximum mass that a passenger can be is known (m). The length of the chain is known (l). Given this information, derive a formula that will tell us the maximum angular velocity before the chain breaks.



Homework Equations



At first, I thought this was easy where T = mgcosθ + mω2rsinθ . But then I realized there were several problems with this firstly that we don't know theta and so can't work out ω when θ is also unknown. But also that using r is incorrect as the true radius that the passenger is revolving round is larger due to the centripetal force making them swing out.

I derived this formula for the necesarry condintions for the passenger being in equilibrium mgsinθ = mω2cosθ(r + lcosθ) . However it doesn't help me much at this stage. The problem is I know that θ and ω are dependent in that if I determine an angular velocity, I will get a corresponding value for θ when the system is in equilibrium.
 
Physics news on Phys.org
T is in fact ##\vec T ##. You can't just add the components.

Draw a free body diagram for the person in the seat and realize that the sum of ##\vec T ## and ##m\vec g## should do nothing else than cause a circular trajectory.
 
Can I ask why you are taking the sine of the centrifugal force?
 
Does anyone how I can draw up and post free body diagrams? It would help me respond to the last two replies with much better clarity.
 
When you go advanced you get the opportunity to upload a graphics file (under attachments). What kind depends on what you have available to make one.
 
Ok well for now I will try to get through this with plain algebra. After correcting and refining my equations, I've come to the conclusion that the angle theta can be no larger than such a value that mg/cosθ could exceed T. In other words the maximum value for theta is θ=〖cos〗^(-1) mg/T .
However, the maximum angle is not what I want, I want to know the maximum angular velocity I can use. A formula I was able to derive is ω = √(gsinθ/(cosθ(r+lsinθ))) . If I substitute θ for 〖cos〗^(-1) mg/T I can determine what the angular velocity should be using all known values (as I don't know what theta is).

I hope this is right though!
 
I think your first attempt was better.

You can solve this in two ways:

1) set up the equations of motion and solve a differential equation
2) use conservation of energy to get the maximum ω
 
Ok here's my attempt at setting up an equation of motion with time.
I going to define the variable x, as the distance across the circumference of the quarter circle the passenger travels as the centripetal force throws them ourwards across this path. The other variable is time t. The angular velocity ω is constant as is r and the length of the chain l.

So, as the carousel swing starts to spin we get a force mω2rcosθ pushing the passenger up. However the force is increasing with time because the radius is increasing as θ increases (but is also decreasing because of cosθ (whether on the whole it is increasing or decreasing I don't know)). So the force now looks like mω2(r+lsinθ)cosθ. However I don't want theta here as it is a third variable, but I can define it terms of x. θ (in radians) = x/l . So it now looks like this mω2(r+lsin(x/l))cos(x/l) . However, there is also another force acting directly against this which is mgsin(x/l) and it increases with time however I don't know at which stage this force is less than the opposing force and then is greater and then is equal, however I do know that these changes happen.

So from here, I have no idea what to do nor can I do anything with conservation of energy.
 
Some notes:
1) You have 2 forces acting on your mass m: T and mg
2) The mω^2r is the centrifugal force which is simply the resultant of T and mgcosθ
3) The radius r does not change with time
4) θ=x/l is an approximation that is valid for small values of θ only ≈60° or less for ≈5% error.
5) Set up the 3 equations of motion: ∑Fx=max, ∑Fy=may, ∑M=Iα. The problem can be simplified if you take the y-direction to be in the same direction as T
6) Your formula looks wrong. It should come out to be a simple looking differential equation at the end. You can solve the differential equation to get ω.
 
  • #10
Sorry I appreciate your time but I can't understand how 3) (x is the circumference and l is a radius) and 4) are true and how my formula is wrong and I can't understand points 2) and 5).
Sorry again.
 
  • #11
That's why a diagram of the situation would help explain better.

I assumed we were talking about a simple pendulum that is free to swing back and forth. I see know you seem to be talking about something else.
 

Similar threads

Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
9
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
25
Views
2K