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- Thread starter entphy
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Rotational kinetic energy is proportional to rotational speed

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Kinetic energy is not necessarily conserved in a closed system. Internal forces can convert kinetic energy to other forms of energy.

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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.

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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