I Energy release - temperature

1. Aug 4, 2016

cptolemy

Hi,

I would like to ask a question, if I may: if I have a body with a calculated Inertial moment at time t0, that in a space of time delta t changes, there's a variation in its kinetic energy, right?

How can I calculate the released temperature in that elapsed time? Is that energy all lost in temperature?

If yes, do I convert only the Joules to Celsius?

Kind regards,

Kepler

2. Aug 5, 2016

BvU

There is such a thing as conservation of angular momentum. If you change $I$, $I\omega$ is constant, so apparently ${1\over 2} I\omega^2$ changes. That means some work has been done. I don't see a direct relation with temperature (negative work can also be converted to spring energy, for example).

Anyway, temperature is not something you can 'release'. But yes, energy can be converted into heat (thermal energy). The change in temperature then depends on specific heat and your 'conversion' looks like $\Delta E = mc_p\Delta T$

3. Aug 5, 2016

jbriggs444

There can be complexities here. We are not told that the object is rigidly rotating. If you take a skater with her arms pulled in close to the body and she then extends the arms then indeed, moment of inertia has increased, rotational kinetic energy given by $\frac{1}{2}I \omega ^2$ has been lost and likely converted into heat.

But if you instead simply chop the skater's arms off, the resulting motion is no longer a rigid rotation, kinetic energy is no longer calculated as $\frac{1}{2} I \omega ^2$. The moment of inertia can still be calculated and has still increased, but there is no change in kinetic energy and no increase in heat.

4. Aug 5, 2016

cptolemy

Good morning,

Thanks for the help. Let's supose a sphere rotating around its axis at a velocity w0. It will have a KE1=1/2*I*w0^2. Supose that in an interval delta t, the axis velocity decreases to w1: KE2 = 1/2*I*w1^2. So delta KE = 1/2*I*(w0^2 - w1^2) in the time interval delta t. Is this correct?

If yes, how do I estimate this variation in the KE in terms of temperature?

Kind regards,

Kepler

5. Aug 5, 2016

Staff: Mentor

There's no good answer without more infomation. Something had to interact with the sphere to carry away the excess angular momentum, but that interaction will also have transferred some kinetic energy from the sphere to the whatever else was involved in the interaction, and that energy will not be available to heat the sphere. Without the details the interaction there's no way of knowing how much of the change in kinetic energy went to heat the sphere and how much went somewhere else.

6. Aug 5, 2016

nasu

And even when (if) you know how much heat went into the sphere, you still need information about the sphere thermal properties and about boundary conditions in order to find a temperature. It may be a temperature distribution rather than a single value.

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