Question

You climb into your seat at the bottom of a Ferris wheel which then starts from rest and has uniform angular acceleration of $\frac{2}{\pi }rad/s^2.$ The Ferris wheel has radius 3.6 m. just as you reach the top the link of your seat breaks? What speed do you hit me ground with?

Answer 1

At the top , let its speed is $u$ and angular speed is $\omega$.

${\omega }^2=2\alpha \theta =2\times \frac{2}{\pi }\times \pi =4$

$⟹\omega =2rad/s$

$u=\omega R=2R$

From conservation of mechanical energy during falling-

$\frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2+mgh$

$⟹v^2=u^2+2g\left(2R\right)$ height it falls is equal to its diameter and hence $h=2R$

$⟹v=\sqrt{u^2+4gR}=\sqrt{4R^2+4gR}=\sqrt{4\times {3.6}^2+4\times 10\times 3.6}$

$⟹v=14m/s$