Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum 
Aug 4th 2016, 05:26 PM

#1  Junior Member
Join Date: Aug 2016
Posts: 4
 Inertial moment variation
Hi,
I would like to ask a question, if I may: if I have a body with a calculated Inertial moment and angualr momentum 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

 
Aug 5th 2016, 04:59 AM

#2  Senior Member
Join Date: Jun 2016 Location: England
Posts: 590

There are 2 ways I can think of to change the (rotational) inertia of an object, 1 is to change its mass, 2 is to change the distribution of mass.
If you change the mass then the system is not closed and the change in mass is changing the energy (not any change in temperature).
If you just move the mass (like a spinning skater pulling in their arms to spin faster) then the kinetic energy stays the same, it is just redistributed differently through the system. So again no change in temperature.

 
Aug 5th 2016, 06:18 AM

#3  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,299

It is possible to have a change in angular or linear momentum cause a change in temperature. In thermodynamics terms the energy applied to a closed system equals the work done by that system on its surroundings plus the increase in its internal energy: dQ = dU+dW. Recall that internal energy is the sum of the system's kinetic energy, potential energy, and heat energy. The question asked here is about decreasing angular and/or linear momentum  if there is no external force applied to the system (i.e. dQ = 0), then any decrease in KE must be matched by any or all of the following: an increase in potential energy, and increase in temperature, and/or work done by the system on its surroundings. (N.B. I am ignoring the possibility of changes in nonmechanical types of energies, such as chemical energy, or atomic energy, or electrical energy here, as these are outside the scope of the question.) Taking it one step further  if we are told that the change in PE is 0 (e.g. the object is not moving uphill), and also are told that the object does no work on its surroundings (i.e. we can neglect external friction and air resistance), then it must be that heat energy (temperature) is increasing by an amount equivalent to the loss in KE. Here's an example: consider the internal friction of a mechanism, such as an gear on a bearing, which has some initial value of angular momentum, and hence KE. Assume the gear set is enclosed in a insulated container such that there is no heat flow through the container walls, and no friction between the gear set and the outside world. As the gear rotation slows due to friction of its bearings inside the container there will be a corresponding temperature rise of the system as its heat energy increases by an amount equal to the loss of KE.
Last edited by ChipB; Aug 5th 2016 at 06:21 AM.

 
Aug 5th 2016, 06:51 AM

#4  Junior Member
Join Date: Aug 2016
Posts: 4

Good morning,
Thanks for the help. Let's supose a sphere rotating around its axis at a angular velocity w0. It will have a KE1=1/2*I*w0^2. Supose that in an interval delta t, the angular 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,
JKepler

 
Aug 5th 2016, 07:24 AM

#5  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,299

Originally Posted by jkepler So delta KE = 1/2*I*(w0^2  w1^2) in the time interval delta t. Is this correct? 
Almost correct. The change in KE is equal to the final KE minus the initial  you have it backwards. You should end up with a negative number.
Originally Posted by jkepler If yes, how do I estimate this variation in the KE in terms of temperature? 
You must first clarify that there are no external forces operating on the sphere, that the sphere does no work against anything external to it, that the sphere is in a vacuum (so that there is no air to conduct heat away), that the sphere is a perfect heat conductor (so that there is no variation in temperature throughout the sphere), and you can ignore any radiation of heat from the sphere into space. Under all those conditions the change in internal energy of the sphere is 0, and hence the increase in heat of the sphere equals the decrease in KE. The increase in temperature corresponds to the decrease in KE according to:
Delta_T x C_p x m + Delta_KE = 0.
where Delta_KE is the change in KE in Joules, C_p is the specific heat of the material(in Joules/KgK, and m is the mass of the sphere in Kg

 
Aug 5th 2016, 11:06 AM

#6  Junior Member
Join Date: Aug 2016
Posts: 4

Originally Posted by ChipB Almost correct. The change in KE is equal to the final KE minus the initial  you have it backwards. You should end up with a negative number. The increase in temperature corresponds to the decrease in KE according to:
Delta_T x C_p x m + Delta_KE = 0.
where Delta_KE is the change in KE in Joules, C_p is the specific heat of the material(in Joules/KgK, and m is the mass of the sphere in Kg 
Hi,
Sorry for the wrong sign... :/
So, Delta_T = Delta_KE / (C_p x m) Celsius?... I thing this doesn't take into acount the delta t time of decreasing angulat velocity... :/
Kind regards,
JKepler

 
Aug 5th 2016, 11:48 AM

#7  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,299

Originally Posted by jkepler So, Delta_T = Delta_KE / (C_p x m) Celsius? 
Strictly speaking Delta T is in Kelvin, but degrees Celsius works too.
Originally Posted by jkepler .. I thing this doesn't take into acount the delta t time of decreasing angulat velocity... 
All that matters is the change of KE  it doesn't matter how long it takes (as long as there is no leaking of heat away from the sphere). Why would the time matter?
By the way, I'm curious if the problem you are working on says anything about the mechanism that causes the sphere to slow its rotation  does it?

 
Aug 5th 2016, 04:46 PM

#8  Junior Member
Join Date: Aug 2016
Posts: 4

Originally Posted by ChipB Strictly speaking Delta T is in Kelvin, but degrees Celsius works too.
All that matters is the change of KE  it doesn't matter how long it takes (as long as there is no leaking of heat away from the sphere). Why would the time matter?
By the way, I'm curious if the problem you are working on says anything about the mechanism that causes the sphere to slow its rotation  does it? 
Hi,
Actually I'm studying an hipothetical case. Supose that our minor companion planet Mercury had a small moon for a while (almost impossible of course). What would be the temperature released if that moon made Mercury slow down its axial rotation in about half an hour in the period of 3 hours?
This is a case study. Care to share your thoughts?
Kind regards,
JKepler

 
Aug 6th 2016, 05:53 AM

#9  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,299

It's a bizarre hypothetical  the interactions between moons and planets can cause some decrease in rotational velocity of the planet through tidal forces, but the effect is small. Our own moon has had a similar effect on Earth  drag from tidal forces has caused Earth's rotational period to slow by about 1.7 milliseconds over the past 100 years. I'm curious how you think a moon could effect Mercury's rotation so significantly? But to answer your question: I think you could calculate the amount of KE lost and as a first order calculation you could use the equations I gave you. You could assume a value of C_p similar to granite, which is about 800 J/KgK. This would ignore factors such as the heat that would be radiated into space, thus cooling the planet's surface, and it would assume a constant change in temperature throughout the entire planet. I suggest you try it, and post back with what your calculation shows.

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