What is the maximum speed for a fairground roundabout with a 20º angle limit?

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Homework Help Overview

The problem involves a fairground roundabout with carriages suspended from a rotating roof by chains. The chains are 3.5m long, and the roundabout's roof has a diameter of 6m. The carriages must maintain a maximum angle of 20º to the vertical during operation. Participants are tasked with determining the maximum speed in revolutions per minute (rpm) at which the roundabout can safely operate.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Some participants discuss whether to include the radius of the circular roof in their calculations, questioning how it affects the radius of the circular path taken by the carriages.
  • There are attempts to derive equations for maximum speed based on gravitational forces and centripetal acceleration, with varying interpretations of the radius involved.
  • One participant suggests using a period calculation based on a derived radius, while another emphasizes the importance of drawing a free body diagram to identify forces acting on the chair.
  • Questions arise regarding the correct application of Newton's laws and the need for a clear understanding of the forces involved.

Discussion Status

The discussion is ongoing, with various interpretations and approaches being explored. Some participants provide guidance on the importance of foundational concepts like free body diagrams and centripetal forces. However, there is no explicit consensus on the correct method or values to use in the calculations.

Contextual Notes

Participants note potential confusion regarding the radius of the circular path and the forces acting on the carriages. There is also mention of a requirement to draw a sketch, which may not have been completed by all participants. The problem's constraints and the necessity for accurate assumptions are under discussion.

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Homework Statement


1 A fairground roundabout has a number of carriages hung from a rotating circular roof by means of chains 3.5m long. The upper ends of the chains are attached to the edge of the circular roof which has a diameter of 6m. During operation, the carriages must not exceed an angle of 20º to the vertical. Draw a good annotated sketch of the roundabout and determine the maximum speed in rev.min-1 at which it can operate.


Homework Equations


I don't know whether to enter the circular roof's radius of 3 m in the equation below, ie add it to the 3.5 m long chain?


The Attempt at a Solution


Without addition of the roofs radius of 3 m:
\sqrt{}(9.81tan20)/(3.5sin20) = 1.727rad.sec
1.727 x 60 / 2\pi = 16.49 rpm

Addition of the radius of 3 m:
\sqrt{}(9.81tan20)/(6.5sin20) = 1.267rad.sec
1.727 x 60 / 2\pi = 12.10 rpm
 
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mucky said:

Homework Statement


1 A fairground roundabout has a number of carriages hung from a rotating circular roof by means of chains 3.5m long. The upper ends of the chains are attached to the edge of the circular roof which has a diameter of 6m. During operation, the carriages must not exceed an angle of 20º to the vertical. Draw a good annotated sketch of the roundabout and determine the maximum speed in rev.min-1 at which it can operate.


Homework Equations


I don't know whether to enter the circular roof's radius of 3 m in the equation below, ie add it to the 3.5 m long chain?
determine the radius of the curve that the chair is rotating in.

The Attempt at a Solution


Without addition of the roofs radius of 3 m:
\sqrt{}(9.81tan20)/(3.5sin20) = 1.727rad.sec
1.727 x 60 / 2\pi = 16.49 rpm
looks like you've combined a couple of steps here, and made an algebra error. Please show step by step how you arrived at this equation.
Addition of the radius of 3 m:
\sqrt{}(9.81tan20)/(6.5sin20) = 1.267rad.sec
1.727 x 60 / 2\pi = 12.10 rpm
Same comment as above, but also, what is the radius of the curve? You don't have the correct value. Did you draw a sketch?
 
Sooooo...
Circular radius: sin20 x 3.5 = 1.197m

Then could I:
2\pi\sqrt{}1.197/9.81tan20 = 3.638s
Because I have the period time: 1/f = 1/3.638 = 0.275 revs/per/sec
0.275 x 60 = 16.49rpm
message back

Formula I received from my tutor was for a similar problem
but can't be used in this one! I think I'm still wrong as I have
used a slightly strange route, could you help me solve it please.
Thanks
 
You just can't look at one 'similar' problem and expect to solve another. You've got to get right down to the basics by drawing a sketch of the problem, identify the forces acting on the chair, draw a good free body diagram of the chair, and apply Newton's laws and your unsderstanding of cenrtipetal forces and centripetal acceleration.
By drawing a sketch, you should be able to note that the radius of the circle in which the chair moves is (3 + 3.5sin20). Now identify the forces acting on the chair in the x and y directions, and apply Newton's laws. The chair does not accelerate in the y direction, but id does accelerate centripetally in the x direction. Are you familiar with free body diagrams and Newton's laws?
 
panthomjay is right.
understand that the centripetal acceleration along the x-axis will be the resultant of the weight of the chair and the tension in the chain. when this ia at the required angle you will be able th calculate Fcentripetal and also tangential speed - do you know the mass involved etc?