Why rotational kinetic energy of a closed system is not conserved?

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A simple rotating system with no external forces acting on it carries a fixed angular momentum and an associated rotational kinetic energy. If the system changes its internal configuration, such as a spinning skater retracting or extending his/her arms, the angular momentum remains constant since no external torque is applied on the system. However, the rotational kinetic energy changes with the configuration. How is this possible when there is no external work done involved? Where does the change of the rotational energy go or come from?
 

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rcgldr
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Even if the work done is internal, that work can result in a change in one form of energy for another. A skater compicates the situation, because chemical energy is consumed when just maintiaining a force without any movement (so no work done).
 
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A simple rotating system with no external forces acting on it carries a fixed angular momentum and an associated rotational kinetic energy. If the system changes its internal configuration, such as a spinning skater retracting or extending his/her arms, the angular momentum remains constant since no external torque is applied on the system. However, the rotational kinetic energy changes with the configuration. How is this possible when there is no external work done involved? Where does the change of the rotational energy go or come from?

Rotational kinetic energy is proportional to rotational speed squared, whereas angular momentum is proportional to rotational speed (^1), so the skater is increasing kinetic energy without increasing angular momentum by increasing their rotational speed. The energy comes from the work done in reducing the moment of inertia of a rotating body (it takes effort to move your arms inward when spinning).
 
  • #4
nasu
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This is not something specific for rotational kinetic energy.
Kinetic energy is not necessarily conserved in a closed system. Internal forces can convert kinetic energy to other forms of energy.
 
  • #5
rcgldr
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As an example, imagine two disks spinning in opposite directions, but connected by a coil, that causes them to cycle back and forth. Stopping for a instant at the end / beginning of each cycle, with kinetic energy zero and spring potential energy at it's max. In the middle of the cycle, kinetic energy is at a max, and sprint potential energy is zero. It's the rotational equivalent of a spring and mass system in a gravity free environment.
 
  • #6
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Thank you everybody for the promptly replies. Yes, the workdone against the centripetal force in the radial direction about the axis of rotation fully accounts for the rotational kinetic energy difference while conserving angular momentum throughout the entire process of internal config change, assuming no external torque is acting on the system.
Mathematically, taking a system of 2 point objects of mass m each attached at the 2 ends of a massless rigid rod, distance 2r apart. The objects are adjustable along the rod. If the system is set to rotate about the center of mass at a constant angular velocity w,
The angular momentum L = 2mr2w, which is a constant without external torque acting on the system.
The centrifugal force at any instant for each object is F = mrw2 = L2/(4mr3).
The workdone to move the 2 objects in synchrony along the rod from position ri to rf with respect to the center of mass are W = -∫_("ri" )^"rf" ▒2Fdr = -∫_("ri" )^"rf" ▒〖2L"2" /((4mr"3" ) ) dr〗 = L"2" /((4m) )[1/((r"f3" ) )-1/((r"i2" ) )]
And the last expression is simply the delta of the rotational kinetic energy for the system moving from position ri to rf , because the rotational kinetic energy at position r is simply 2mr2w2/2= L2/(4mr2).

*ps. apology I think the format for the mathematical expressions simply do not come out right, it doesn't work by pasting from MS Words. Please refer to the attachment.
 

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