Question

An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the end of 1.14 m massless chains. When the system rotates, the chains make an angle of 16.2 degrees with the vertical. The acceleration of gravity is 9.8 m/s^2.1. What is the speed of each seat? I fixed my answer and got 3.57 m/s.2. If a child of mass 26.2 kg sits in a seat, what is the tension in the chain (for the same angle)?

Answer 1

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

$r = \frac{8.26}{2} = 4.13m$

And the distance 'd' is

$d = lsin\theta$

$d = 1.14 sin 16.2\°$

$d = 0.318m$

Through the free-body diagram the tension components are given by

$Tcos\theta = mg$

$Tsin\theta = \frac{mv^2}{R}$

Here we can watch that,

$R = r+d$

Dividing both expression we have that,

$tan\theta = \frac{v^2}{Rg}$

Replacing the values,

$tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}$

$v = 4.83371m/s$

PART B) Using the vertical component we can find the tension,

$Tcos\theta = mg$

$T = \frac{mg}{cos\theta}$

$T = \frac{(10+26.2)(9.8)}{cos(16.2)}$

$T = 369.42N$