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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. 1-85

 [1-86]

However, power is the rate of doing work, and we have

 [1-87]

where dω/dt is angular acceleration in radians per sec2.

Suppose that, to develop the power expressed by Eq. 1-87, an external force F is applied to a point at a distance R from the center; then from p = Fv it follows that

 [1-88]

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. 1-88 becomes p = ωT, and we have

 [1-89]

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 RM. Then if the moment of inertia of the flywheel is I, we have

 [1-90]

the solution of which is

 [1-91]

Note that this expression for angular velocity has the same form as that for a current in an inductive circuit with constant applied d-c voltage, as in Eq. 1-40. The moment of resistance RM corresponds to the electrical resistance of the circuit, and the moment of inertia I corresponds to the self-inductance L. Furthermore, the energy stored in the flywheel is expressed by Iω2/2, and the energy stored in the self-inductance is expressed by Li2/2.

Last Update: 2011-01-04