Question

An amusement park ride consists of a rotating
circular platform 9.15 m in diameter from
which 10 kg seats are suspended at the end
of 3.45 m massless chains. When the system
rotates, the chains make an angle of 42.6◦ with
the vertical.
The acceleration of gravity is 9.8 m/s2.
What is the speed of each seat?
Answer in units of m/s.

Answer 1

Answer:

Let T be the tension in the chain,

Then,

In the vertical direction,

$Tcos(\theta) = mg \to (1)$

Where m - mass of a seat

g - acceleration of gravity

In the horizontal direction,

$Tsin(\theta) = \frac{mv^2}{r} \to (2)$

where, v - speed of a seat

According to the figure,

$r = lsin(\theta) + \frac{d}{2}$

$r = 3.45sin(42.6) + \frac{9.15}{2} = 6.91m$

dividing equation (2) by (1)

we can get,

$tan(\theta) = \frac{v^2}{rg}$

Therefore, the speed of a seat,

$v = \sqrt{rgtan(\theta)}$

$v = \sqrt{6.91*9.8*tan(42.6)}$

$v = 7.89 ms^{-1}$