An amusement park ride consists of a rotating circular platform 8.00 m in diameter from which 9.5 kg seats are suspended at the end of 2.50 m massless chains. When the system rotates, the chains make an angle θ = 31.0° with the vertical. Assume the average mass of a child is 40.0 kg.
(b) Find the tension in the chain.
(M+m) •a=T•sinθ
(M+m)g= T•cosθ
a/g= T•sinθ/ T•cosθ=tanθ
a=g• tanθ
T=(M+m) •a/ sinθ=(M+m) • g• tanθ/ sinθ
To find the tension in the chain, we can use the concept of forces in equilibrium.
Let's consider one of the seats and draw a free body diagram for it.
First, we have the weight of the seat acting downwards, which can be calculated as:
Weight = mass * gravitational acceleration
= 9.5 kg * 9.8 m/s^2
= 93.1 N
Next, we have the tension in the chain, which is acting upwards at an angle of 31.0° with the vertical. This tension can be resolved into two components: one in the vertical direction and one in the horizontal direction.
The vertical component of the tension balances the weight of the seat, so we can calculate it as:
Vertical component of tension = Weight * cos(θ)
= 93.1 N * cos(31.0°)
The horizontal component of the tension doesn't contribute to the equilibrium, so it is not significant for this problem.
Therefore, the tension in the chain is equal to the vertical component of the tension, which can be calculated using the equation above.