- #1

- 412

- 2

Got this question somewhere: A boy is standing on a rotating platform. The system has a kinetic energy 'K'. Now, the boy stretches out is hand such that the Moment of Inertia of the system doubles. What is the Kinetic Energy of the system?

So.. i first applied conservation of angular momentum,

[tex]

I\omega_1 = 2I\omega_2

[/tex]

(here, [itex]\omega_1[/itex] and [itex]\omega_2[/itex] are the angular velocities before & after he stretches his hands).

So, we have:

[tex]

\omega_2 = \frac{\omega_1}{2}

[/tex]

Also,

[tex]

K = \frac{1}{2}{\omega_1}^2

[/tex]

and the new kinetic energy is:

[tex]

K_2= \frac{1}{2}{\omega_2}^2 = \frac{1}{8}{\omega_1}^2

[/tex]

So, the energy difference is:

[tex]

K - K_2 = \frac{3}{8}{\omega_1}^2

[/tex]

So.. where exactly does this energy go? By COE, this needs to be translated to the potential energy.. but does something called 'Rotational Potential Energy' even exist? How do i calculate it if it does?

So.. i first applied conservation of angular momentum,

[tex]

I\omega_1 = 2I\omega_2

[/tex]

(here, [itex]\omega_1[/itex] and [itex]\omega_2[/itex] are the angular velocities before & after he stretches his hands).

So, we have:

[tex]

\omega_2 = \frac{\omega_1}{2}

[/tex]

Also,

[tex]

K = \frac{1}{2}{\omega_1}^2

[/tex]

and the new kinetic energy is:

[tex]

K_2= \frac{1}{2}{\omega_2}^2 = \frac{1}{8}{\omega_1}^2

[/tex]

So, the energy difference is:

[tex]

K - K_2 = \frac{3}{8}{\omega_1}^2

[/tex]

So.. where exactly does this energy go? By COE, this needs to be translated to the potential energy.. but does something called 'Rotational Potential Energy' even exist? How do i calculate it if it does?

Last edited: