# Where does the K.E go? [RotMech]

• rohanprabhu
In summary, a boy standing on a rotating platform stretches his hands, causing the moment of inertia of the system to double. The kinetic energy of the system is then calculated using the conservation of angular momentum and is found to be half of the original kinetic energy. The energy difference is then discussed and it is determined that mechanical energy is not conserved due to the work done by the boy in moving his hands. The concept of non-conservative forces is introduced and it is explained that the energy consumed by the boy goes into various forms including heat. The work energy theorem is also mentioned and it is concluded that the work done in this case is zero. It is clarified that when the boy brings his hands back together, the system's kinetic energy increases due
rohanprabhu
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,

$$I\omega_1 = 2I\omega_2$$

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

So, we have:

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

Also,

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

and the new kinetic energy is:

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

So, the energy difference is:

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

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?

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erm the kinetic energy of a rotation is 1/2I*w^2

rohanprabhu said:
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,

$$I\omega_1 = 2I\omega_2$$

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

So, we have:

$$\omega_2 = \frac{\omega_1}{2}$$
OK. Let's at least imagine that he's holding weights in his hands or something to give him a bit of rotational inertia to play with.

Also,

$$K = \frac{1}{2}{\omega_1}^2$$
That should be:
$$K_1 = \frac{1}{2}I{\omega_1}^2$$
and the new kinetic energy is:

$$K_2= \frac{1}{2}{\omega_2}^2 = \frac{1}{8}{\omega_1}^2$$
That should be:
$$K_2 = \frac{1}{2}I_2{\omega_2}^2 = \frac{1}{2}( \frac{1}{2}I{\omega_1}^2)$$

So, the energy difference is:

$$K - K_2 = \frac{3}{8}{\omega_1}^2$$
$$K_2 = \frac{1}{2} K_1$$
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?
Mechanical energy is not conserved. (Consider the work he must do as he extends his hands.)

Doc Al said:
OK. Let's at least imagine that he's holding weights in his hands or something to give him a bit of rotational inertia to play with.

That should be:
$$K_2 = \frac{1}{2}I_2{\omega_2}^2 = \frac{1}{2}( \frac{1}{2}I{\omega_1}^2)$$

$$K_2 = \frac{1}{2} K_1$$

Mechanical energy is not conserved. (Consider the work he must do as he extends his hands.)

umm.. right.. i made that slip there.. it's not 1/4th but half of the K.E. but that apart, you mean to say that the work done by the boy to move his hands apart equals the difference [taking the general assumptions of no air friction and such] of the 2 kinetic energies? So, if he brings his hands back together, the work done will translate into kinetic energy again and bring it back to the previous kinetic energy 'K'.

But here, the boy has done work.. but still the system remains unchanged. Where does the energy consumed by the boy go?

By Work energy theorem, the change in Kinetic energy is given by the total work done. Which means, the work done in this case is zero??

Could somebody please clarify this a little bit..

well, in that case, i guess the boy applied two non-conservative forces, making the system to apparently(literally) conserve itself...

Whoever, energy was lost to the boy make that movements. In general, Men system are non conservative, as it was a great process to make him move(or being alive), heat is only one of the K.E. lost...

rohanprabhu said:
umm.. right.. i made that slip there.. it's not 1/4th but half of the K.E. but that apart, you mean to say that the work done by the boy to move his hands apart equals the difference [taking the general assumptions of no air friction and such] of the 2 kinetic energies? So, if he brings his hands back together, the work done will translate into kinetic energy again and bring it back to the previous kinetic energy 'K'.

But here, the boy has done work.. but still the system remains unchanged. Where does the energy consumed by the boy go?

By Work energy theorem, the change in Kinetic energy is given by the total work done. Which means, the work done in this case is zero??

Could somebody please clarify this a little bit..
When he brings his hands back, he makes mechanical work to the system, which then increases its kinetic energy; when he let's his hands moving apart, the work he makes on the system is negative, that is, the system makes work on him.

Think of a stretched spring connected with the two masses: when the spring is released and can contract making the massess approach, its previous elastic potential energy is converted into system's kinetic energy; if you instead had a non-stretched spring, then centrifugal force makes it stretch and system's kinetic energy is converted into spring potential energy.
(There are no tricks! Energy is neither created nor destroyed!)

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## 1. Where does the kinetic energy go when an object is rotating?

When an object is rotating, the kinetic energy is stored in the object's rotational motion. This energy is known as rotational kinetic energy and is directly related to the object's angular velocity and moment of inertia.

## 2. Does the kinetic energy decrease as an object rotates?

No, the kinetic energy does not decrease as an object rotates. In fact, the kinetic energy will remain constant as long as there is no external force acting on the object to slow it down.

## 3. How does friction affect the kinetic energy of a rotating object?

Friction can cause the kinetic energy of a rotating object to decrease over time. This is because friction acts as an external force that opposes the object's rotational motion, causing it to slow down and lose kinetic energy.

## 4. Can kinetic energy be converted into rotational energy?

Yes, kinetic energy can be converted into rotational energy. This can occur when an external force acts on an object and causes it to rotate, such as when a person pushes a merry-go-round to set it in motion.

## 5. Is rotational kinetic energy different from linear kinetic energy?

Yes, rotational kinetic energy is different from linear kinetic energy. While linear kinetic energy is associated with an object's linear motion, rotational kinetic energy is associated with an object's rotational motion. They are calculated using different equations and have different units of measurement.

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