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Belt conveyor (2.2 m/s) load inertia and torque

Short answer

Mass on the belt is 2525.3 kg; load inertia referred to the motor shaft is Jl = M(v/w)^2 = 0.495 kg-m2. Adding motor inertia gives 0.595 kg-m2, acceleration is 19.63 rad/s2, so motor torque is about 11.7 N-m.

A horizontal conveyer belt moving at uniform speed of 2.2 m/s transport material at the rate of 200 tonnes/hour. Belt is 100 m long and driven by a motor at 1500 rpm. i) Determine load inertia referred to motor shaft. ii) Calculate torque that motor should develop to accelerate the belt from standstill to full speed in 8 second. Moment of inertia of motor is 0.1 kg-m².

Assumptions: the belt's own mass and friction are neglected (not given); the whole 100 m length carries material.

Mass of material on the belt

rate=200×10003600=55.56 kg/stbelt=1002.2=45.45 sM=55.56×45.45=2525.3 kg\begin{aligned} \text{rate} &= \frac{200\times1000}{3600} = 55.56\ \text{kg/s} \\ t_{belt} &= \frac{100}{2.2} = 45.45\ \text{s} \\ M &= 55.56 \times 45.45 = 2525.3\ \text{kg} \end{aligned}

Motor speed

ωm=2π×150060=157.08 rad/s\omega_m = \frac{2\pi\times1500}{60} = 157.08\ \text{rad/s}

i) Load inertia referred to motor shaft

Equating kinetic energies, 12Jlωm2=12Mv2\tfrac12 J_l\omega_m^2 = \tfrac12Mv^2:

Jl=M(vωm)2=2525.3×(2.2157.08)2=2525.3×1.962×10−4=0.495 kg-m2\begin{aligned} J_l &= M\left(\frac{v}{\omega_m}\right)^2 = 2525.3\times\left(\frac{2.2}{157.08}\right)^2 \\ &= 2525.3\times1.962\times10^{-4} = 0.495\ \text{kg-m}^2 \end{aligned}

ii) Torque to accelerate in 8 s

J=Jm+Jl=0.1+0.495=0.595 kg-m2J = J_m + J_l = 0.1 + 0.495 = 0.595\ \text{kg-m}^2 dωmdt=157.088=19.63 rad/s2\frac{d\omega_m}{dt} = \frac{157.08}{8} = 19.63\ \text{rad/s}^2 T=Jdωmdt=0.595×19.63=11.69 N-mT = J\frac{d\omega_m}{dt} = 0.595\times19.63 = 11.69\ \text{N-m}

The conveyor is horizontal and friction is neglected, so no steady load torque is added.

Answer: (i) Jl=0.495J_l = 0.495 kg-m²; (ii) accelerating torque ≈11.7\approx 11.7 N-m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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