Winch lifting 500 kg: motor torque and inertia

Data: $m = 500$ kg, $v = 1.5$ m/s, $N = 1000$ rpm, $J_m = 0.5$ kg-m², $J_{winch} = 0.3$ kg-m² (taken at motor speed), no-load (friction) torque $= 100$ N-m at 1000 rpm.

$$\omega_m = \frac{2\pi\times1000}{60} = 104.72\ \text{rad/s}$$

Motor torque

Torque to lift the weight, from power balance $T\omega_m = Fv$:

$$\begin{aligned} T_w &= \frac{mgv}{\omega_m} = \frac{500\times9.81\times1.5}{104.72} \\ &= \frac{7357.5}{104.72} = 70.26\ \text{N-m} \end{aligned}$$

Without the weight the motor must still supply 100 N-m (friction and other losses at 1000 rpm). At uniform speed there is no accelerating torque, so

$$T = 100 + 70.26 = 170.26\ \text{N-m}$$

Equivalent moment of inertia at motor shaft

The weight moving at $v$ has kinetic energy $\tfrac12mv^2$, referred to the motor shaft as $J_w = m(v/\omega_m)^2$:

$$\begin{aligned} J_w &= 500\times\left(\frac{1.5}{104.72}\right)^2 = 0.1026\ \text{kg-m}^2 \\ J &= J_m + J_{winch} + J_w = 0.5 + 0.3 + 0.1026 = 0.9026\ \text{kg-m}^2 \end{aligned}$$

Answer: motor torque $\approx 170.3$ N-m; equivalent moment of inertia $\approx 0.903$ kg-m².