Assume that the energy source of radiation of the Crab Nebula pulsar is tapped from its rotational kinetic energy. Does the period of the pulsar increase, decrease, or stay the same?
KE=Iω²/2 =I•4π²/2T²
T=2 π•sqrt{I/2•KE}
If KE↓ => T↑
To determine whether the period of the pulsar (the time it takes to complete one rotation) would increase, decrease, or stay the same if its rotational kinetic energy fueled its radiation, we need to analyze the conservation of energy.
The rotational kinetic energy (E_rot) of an object with moment of inertia (I) rotating at an angular velocity (ω) can be expressed as E_rot = (1/2)Iω^2.
In this scenario, if the Crab Nebula pulsar is emitting radiation by converting its rotational kinetic energy, some of the pulsar's rotational kinetic energy would be continuously consumed. As energy is being lost, it affects the angular velocity of the pulsar.
According to the conservation of angular momentum, the product of the moment of inertia and the angular velocity must remain constant if no external torque acts on the system. Mathematically, Iω = constant.
If we assume that the moment of inertia remains constant, then a decrease in the rotational kinetic energy (due to energy being consumed by radiation) would cause a decrease in the angular velocity. Consequently, the period of the pulsar would increase since the time taken for one rotation is inversely proportional to the angular velocity.
Therefore, in this scenario, if the energy source of radiation for the Crab Nebula pulsar is tapped from its rotational kinetic energy, the period of the pulsar would increase.