How Do You Calculate Speed and Tension in a Swinging Fairground Ride?

[bob 911]

There is a fairground ride where chairs connected to chains swing around. (See attachment for diagram).

The length of the supporting chains is 3m
The length of the top beam to the centre is 2m
The supporting chains make an angle of 25 degrees to the vertical.

A) What is the speed of rotation.
B) What is the tension in the chain when a person of 65kg is on at this speed.

My Solution

A
I equated the centripetal force to F1Sin(Theta).
I equated the weight to F1Cos(Theta).
I simultaneously divided the two, so I had Tan(Theta)= (MV^2/R) / (MG)
Rearranging and replacing v with the w equivalent I ended up with
W= square root of ((gtan25)/(r))

Is this the correct method?

B

I am unsure whether the speed itself plays a part here.
Instead do you not just work out the component of the weight acting in the direction of the swing at that angle. This component is equal to the tension that is required.

Or do you have to use the speed to calculate the tension?

Thanks in Advance!

 

[anathema]

A) Your method is correct. To find the speed of rotation, you can use the equation v = √(g * tanθ * r), where g is the acceleration due to gravity, θ is the angle of the chains, and r is the length of the chains. In this case, g = 9.8 m/s², θ = 25 degrees, and r = 3m. So, the speed of rotation would be v = √(9.8 * tan25 * 3) = 5.81 m/s.

B) You are correct in your thinking that the tension in the chain will be equal to the component of the weight acting in the direction of the swing at that angle. This component can be calculated using the formula T = mg * sinθ, where T is the tension, m is the mass of the person (65kg), g is the acceleration due to gravity (9.8 m/s²), and θ is the angle of the chains (25 degrees). So, the tension in the chain when a person of 65kg is on the ride at this speed would be T = (65 * 9.8) * sin25 = 162.44 N.