Question

An amusement park ride consists of a rotating circular platform 11.4 m in diameter from which 10 kg seats are suspended at the end of 1.54 m massless chains. When the system rotates, the chains make an angle of 32.3 ◦ with the vertical. What is the speed of each seat?

Answer 1

Explanation:

Formula for horizontal equilibrium is as follows.

           $T sin (\theta) = m \times \frac{V^{2}}{r}$    

Now, we will calculate the radius as follows.

        r = $\frac{D}{2}$

           = $\frac{11.4 m}{2}$

           = 5.7 m

Also,   $mg tan (\theta) = m \times \frac{V^{2}}{r}$

but from the figure r = $L sin(\theta) + 5.7 m$

So,   r = $(1.54 \times sin (32.3) + 5.7)$

          = $1.54 \times 0.77 \times 5.7$

          = 6.75 m

Now, we will calculate the velocity as follows.

            $V^{2} = g \times tan (\theta) \times r$

                   V  = $\sqrt(9.81 \times tan (32.3) \times 6.75)$

                     = 8.98 m/s

Thus, we can conclude that the speed of each seat is 8.98 m/s.