Belt conveyor load inertia and accelerating torque

Assumptions: the belt's own mass and friction are not given, so only the mass of material on the belt is considered; the belt is fully loaded along its 200 m length.

Mass of material on the belt

Material rate $= \dfrac{100 \times 1000}{3600} = 27.78$ kg/s. Time for material to travel the belt $= \dfrac{200}{1.2} = 166.67$ s.

$$M = 27.78 \times 166.67 = 4629.6\ \text{kg}$$

Motor speed

$$\omega_m = \frac{2\pi \times 1200}{60} = 125.66\ \text{rad/s}$$

(i) Load inertia referred to motor shaft

Equating kinetic energies, $\tfrac12 J_l\omega_m^2 = \tfrac12 Mv^2$:

$$\begin{aligned} J_l &= M\left(\frac{v}{\omega_m}\right)^2 = 4629.6 \times \left(\frac{1.2}{125.66}\right)^2 \\ &= 4629.6 \times 9.119\times10^{-5} = 0.422\ \text{kg-m}^2 \end{aligned}$$

(ii) Accelerating torque

Total inertia referred to the motor shaft:

$$J = J_m + J_l = 0.1 + 0.422 = 0.522\ \text{kg-m}^2$$

Uniform acceleration from rest to full speed in 8 s:

$$\frac{d\omega_m}{dt} = \frac{125.66}{8} = 15.71\ \text{rad/s}^2$$ $$T = J\frac{d\omega_m}{dt} = 0.522 \times 15.71 = 8.20\ \text{N-m}$$

(As the conveyor is horizontal and friction is neglected, there is no steady load torque; with friction, its torque would be added.)

Answer: (i) load inertia referred to motor shaft $= 0.422$ kg-m²; (ii) motor torque to accelerate in 8 s $\approx 8.20$ N-m.