# Centrpetal force query

## 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

$$\sqrt{}(9.81tan20)/(3.5sin20)$$ = 1.727rad.sec
1.727 x 60 / 2$$\pi$$ = 16.49 rpm

$$\sqrt{}(9.81tan20)/(6.5sin20)$$ = 1.267rad.sec
1.727 x 60 / 2$$\pi$$ = 12.10 rpm

Last edited:

PhanthomJay
Homework Helper
Gold Member

## 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

$$\sqrt{}(9.81tan20)/(3.5sin20)$$ = 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.
$$\sqrt{}(9.81tan20)/(6.5sin20)$$ = 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 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