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Centrpetal force query

  • Thread starter mucky
  • Start date
2
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1. 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.


2. 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?


3. The Attempt at a Solution
Without addition of the roofs radius of 3 m:
[tex]\sqrt{}(9.81tan20)/(3.5sin20)[/tex] = 1.727rad.sec
1.727 x 60 / 2[tex]\pi[/tex] = 16.49 rpm

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

Answers and Replies

PhanthomJay
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1. 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.


2. 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.

3. The Attempt at a Solution
Without addition of the roofs radius of 3 m:
[tex]\sqrt{}(9.81tan20)/(3.5sin20)[/tex] = 1.727rad.sec
1.727 x 60 / 2[tex]\pi[/tex] = 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:
[tex]\sqrt{}(9.81tan20)/(6.5sin20)[/tex] = 1.267rad.sec
1.727 x 60 / 2[tex]\pi[/tex] = 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?
 
2
0
Sooooo...
Circular radius: sin20 x 3.5 = 1.197m

Then could I:
2[tex]\pi[/tex][tex]\sqrt{}1.197/9.81tan20[/tex] = 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 recieved 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
 
PhanthomJay
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Homework Helper
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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?
 
16
0
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?
 

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