the net torque TJ = Tem − TL and the equivalent combined inertia is Jeq = JM + JL.
In Fig. 2-7a, each structure has the same inertia as the cylinder in Example 2-2. The load torque TL is negligible. Calculate the required electromagnetic torque, if the speed is to increase linearly from rest to 1800 rpm in 5 s.
Solution
Using the results of Example 2-2, the combined inertia of the system is
The angular acceleration is
Therefore, from Eq. (2-23),
Fig. 2-7 Motor and load torque interaction with a rigid coupling.
Equation (2-23) shows that the net torque is the quantity that causes acceleration, which in turn leads to changes in speed and position. Integrating the acceleration α(t) with respect to time,
where ωm(0) is the speed at t = 0 and τ is a variable of integration. Further integrating ωm(t) in Eq. (2-24) with respect to time yields
where θ(0) is the position at t = 0 and τ is again a variable of integration. Equations (2-23) through (2-25) indicate that torque is the fundamental variable for controlling speed and position. Equations (2-23) through (2-25) can be represented in a block‐diagram form, as shown in Fig. 2-6b.
EXAMPLE 2‐4
Consider that the rotating system shown in Fig. 2-7a, with the combined inertia Jeq = 2 × 0.029 = 0.058 kg ⋅ m2, is required to have the angular speed profile shown in Fig. 2-1b. The load torque is zero. Calculate and plot, as functions of time, the electromagnetic torque required from the motor, and the change in position.
Solution
In the plot of Fig. 2-2b, the magnitude of the acceleration and the deceleration is 100 rad/s2. During the intervals of acceleration and deceleration, since TL = 0,
as shown in Fig. 2-8.
During intervals with a constant speed, no torque is required. Since the position θ is the time‐integral of speed, the resulting change of position (assuming that the initial position is zero) is also plotted in Fig. 2-8.
Fig. 2-8 Speed, torque, and angle variations with time.
In a rotational system shown in Fig. 2-9, if a net torque T causes the cylinder to rotate by a differential angle dθ, the differential work done is
(2-26)
Fig. 2-9 Torque, work, and power.
If this differential rotation takes place in a differential time dt, the power can be expressed as
where ωm = dθ/dt is the angular speed of rotation. Substituting for T from Eq. (2-21) into Eq. (2-27),
Integrating both sides of Eq. (2-28) with respect to time, assuming that the speed ωm and the kinetic energy W at the time t = 0 are both zero, the kinetic energy stored in the rotating mass of inertia J is
This stored kinetic energy can be recovered by making the power p(t) reverse direction, that is, by making p(t) negative.
