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Power and Torque
The work done in accelerating a rotating body from rest to an angular velocity of ω radians per sec is equal to the kinetic energy of rotation, hence, from Eq. 185
However, power is the rate of doing work, and we have
where dω/dt is angular acceleration in radians per sec^{2}. Suppose that, to develop the power expressed by Eq. 187, an external force F is applied to a point at a distance R from the center; then from p = Fv it follows that
because v = Rω. The product FR is called torque and in the rationalized MKS system is expressed in newton meters. Thus, if the torque is expressed by T = FR, Eq. 188 becomes p = ωT, and we have
the torque required to accelerate a rotating mass. This also expresses the torque developed by a decelerating mass. Suppose that a constant torque T is applied to a flywheel starting from rest and that the moment of resistance of the bearings and other moving parts can be represented by R_{M}. Then if the moment of inertia of the flywheel is I, we have
the solution of which is
Note that this expression for angular velocity has the same form as that for a current in an inductive circuit with constant applied dc voltage, as in Eq. 140. The moment of resistance R_{M} corresponds to the electrical resistance of the circuit, and the moment of inertia I corresponds to the selfinductance L. Furthermore, the energy stored in the flywheel is expressed by Iω^{2}/2, and the energy stored in the selfinductance is expressed by Li^{2}/2.


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