---
title: "Belt conveyor (2.2 m/s) load inertia and torque"
subject: "Utilization of Electrical Energy"
chapter: 2
key: "semester_7/utilization_of_electrical_energy:belt-conveyor-2-2ms-inertia-torque"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/belt-conveyor-2-2ms-inertia-torque
site: BCT Engineering
---

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

Utilization of Electrical Energy · Chapter 2: Electric Drive System

- Asked in: 2079 Bhadra, 4+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/belt-conveyor-2-2ms-inertia-torque
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-2

## Question

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.

## 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.

## Answer

**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**

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

**Motor speed**

$$
\omega_m = \frac{2\pi\times1500}{60} = 157.08\ \text{rad/s}
$$

#### i) Load inertia referred to motor shaft

Equating kinetic energies, $\tfrac12 J_l\omega_m^2 = \tfrac12Mv^2$:

$$
\begin{aligned}
J_l &= M\left(\frac{v}{\omega_m}\right)^2 = 2525.3\times\left(\frac{2.2}{157.08}\right)^2 \\
&= 2525.3\times1.962\times10^{-4} = 0.495\ \text{kg-m}^2
\end{aligned}
$$

#### ii) Torque to accelerate in 8 s

$$
J = J_m + J_l = 0.1 + 0.495 = 0.595\ \text{kg-m}^2
$$

$$
\frac{d\omega_m}{dt} = \frac{157.08}{8} = 19.63\ \text{rad/s}^2
$$

$$
T = J\frac{d\omega_m}{dt} = 0.595\times19.63 = 11.69\ \text{N-m}
$$

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

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the mass of material on the belt?
   A. About 2525 kg (55.56 kg/s x 45.45 s)
   B. About 200 000 kg
   C. About 55.6 kg
   D. About 100 kg
   Answer: A. Rate = 200000/3600 = 55.56 kg/s; travel time 100/2.2 = 45.45 s; M = 2525 kg. (Key point: Mass on belt)

2. What is the load inertia referred to the motor shaft?
   A. 0.495 kg-m^2, from M(v/omega)^2
   B. 49.5 kg-m^2
   C. 0.0495 kg-m^2
   D. 2525 kg-m^2
   Answer: A. J_l = 2525.3 x (2.2/157.08)^2 = 0.495 kg-m^2. (Key point: Referred inertia)

3. What total inertia is used for acceleration?
   A. 0.495
   B. 0.1
   C. 0.595 (motor + load)
   D. 0.395
   Answer: C. J = 0.1 + 0.495 = 0.595 kg-m^2. (Key point: Total inertia)

4. What accelerating torque is needed in 8 s?
   A. About 11.7 N-m
   B. About 117 N-m
   C. About 1.17 N-m
   D. About 19.6 N-m
   Answer: A. T = 0.595 x (157.08/8) = 0.595 x 19.63 = 11.7 N-m. (Key point: Torque)

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