Question

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

Answer 1

Use the suitable equation from angular kinematics and find the value of angular velocity, then use the equation of relation between angular and linear velocity, from there we will get the value of initial velocity. Finally use the concept of conservation of mechanical energy and find the speed when we hit the ground.

Complete step by step answer:
At the top of the Ferris wheel let its speed be $u$ and angular speed be $\omega$. Using the equation from angular kinematics, we get
${\omega ^2} = 2\alpha \theta$
Where $\alpha$ is the angular acceleration which remains constant over the time interval.
$\theta$ is the final angle.
$\Rightarrow 2 \times \dfrac{2}{\pi } \times \pi$
$\Rightarrow \omega = 2$
Using the relationship between angular velocity and linear velocity
$u = \omega r$
Where $u$ is the linear velocity and $\omega$ is the angular velocity $r$ is the radius.
$\Rightarrow u = 2R$

From conservation of mechanical energy during falling, mechanical energy is equal to the sum of kinetic energy and potential energy.
$\dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{u^2} + mgh$
$\Rightarrow {v^2} = {u^2} + 2g(2R)$
As we are falling from the top to bottom therefore the diameter of ferris wheel will be equal to the height we fall so $h = 2R$
$\Rightarrow v = \sqrt {{u^2} + 4gR}$
$\Rightarrow v = \sqrt {4{R^2} + 4gR}$
$\Rightarrow v = \sqrt {4 \times {{(3.6)}^2} + 4 \times 10 \times 3.6}$
$\therefore v = 14\dfrac{m}{s}$

Hence, the speed we hit the ground with is $14\dfrac{m}{s}$.

Note: From the relation between angular velocity and linear velocity we get that larger the linear velocity larger be the angular velocity . Also if A larger-radius wheel revolves at the same angular velocity it will cause the body to move at a faster linear speed.