An amusement park ride consists of a rotating circular platform 8.07 m in diameter from which 10 kg seats are suspended at the end of 2.87 m massless chains.

Question

An amusement park ride consists of a rotating circular platform 8.07 m in diameter from which 10 kg seats are suspended at the end of 2.87 m massless chains. When the system rotates, the chains make an angle of 37.3 ◦ with the vertical. The acceleration of gravity is 9.8 m/s 2 .

1) What is the speed of each seat? Answer in units of m/s.

2) If a child of mass 45.3 kg sits in a seat, what is the tension in the chain (for the same angle)? Answer in units of N.

Answer 1

You can first find the radius, r of the motion using

r = radius of the platform + horizontal swing distance

= diameter/2 + Lsinθ = 5.8 m

where L is the length of the massless chain and θ is the displaced angle

a)

If you draw the forces (FBD) acting on the system and use Newton's second law of motion to find the x - and y-component of the forces, you will get

tanθ = v²/rg ==> v = √(rgtanθ), where v is the speed of each each, with g =9.8m/s² . I trust that you can now find v but plugging in the known r, g and θ values.

b) also, from the x- and y-components of the forces, you have

Tcosθ = mg = T = mg/cosθ, where T = tension in the chain and m = 10kg + 45.3kg, and θ = 37.3°