An amusement park ride consists of a rotating circular platform 8.03 m in diameter from which 10 kg seats are suspended at the end of 3.76 m massless chains. When the system rotates, the chains make an angle of 28.6 ◦ with the vertical. The acceleration of gravity is 9.8 m/s 2 .
1. What is the speed of each seat?
2. If a child of mass 26.2 kg sits in a seat, what is the tension in the chain (for the same angle)?

a) 5.57 m/s

b) 403 N

Explanation:

We will find the radius, r of the motion using the formula

r = radius of the platform + horizontal swing distance

r = diameter/2 + Lsinθ, where L is the length of the massless chain and θ is the displaced angle

r = 8.03/2 + 3.76 sin28.6

r = 4.015 + 1.8

r = 5.82 m

a) To get the speed, we use the formula

tanθ = v2/rg, if we make v subject of formula, we have

v = √(rgtanθ), where

v = speed of the seats

g = 9.8m/s²

v = √(5.82 * 9.8 * tan28.6)

v = √(5.82 * 9.8 * 0.545)

v = √31.08

v = 5.57 m/s

b) To find the tension, we use the formula

Tcosθ = mg, making T subject of formula, we have

T = mg/cosθ, where T = tension in the chain and m = 10 kg + 26.2 kg = 36.2 kg and θ = 28.6°

T = 36.2 * 9.8 / cos 28.6

T = 354.76 / 0.88

T = 403 N

friction

explanation:

they us to move forward and backwards