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

# Belt conveyor load inertia and accelerating torque

Utilization of Electrical Energy · Chapter 2: Electric Drive System

- Asked in: 2081 Bhadra; 2070 Chaitra (asked 2 times)
- 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-inertia-accelerating-torque
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-2

## Question

A horizontal belt conveyor moving at a uniform speed of 1.2 m/s transports material at 100 tons/hour. The Belt is 200 m long and is driven by a motor at 1200 rpm. (i) Find load inertia referred to motor shaft, (ii) Find torque of motor to accelerate the belt from standstill to full speed in 8 seconds. Moment of inertia of motor is 0.1 kgm².

## Short answer

Mass on belt is 4629.6 kg; load inertia referred to motor shaft is Jl = M(v/w)^2 = 0.422 kg-m2. Total inertia is 0.522 kg-m2 and acceleration is 15.71 rad/s2, so accelerating torque is 8.20 N-m.

## Answer

**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.

## Check yourself

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

1. How is the mass of material on the belt found?
   A. Material rate x belt travel time (length/speed)
   B. Material rate x motor speed
   C. Belt length x rpm
   D. Material rate / belt length
   Answer: A. M = 27.78 kg/s x (200/1.2 s) = 4629.6 kg. (Key point: Mass on belt)

2. What equation gives the load inertia referred to the motor shaft?
   A. J_l = M v / omega_m
   B. J_l = M (v/omega_m)^2, from equating kinetic energies
   C. J_l = M v^2 omega_m
   D. J_l = M / omega_m
   Answer: B. Equating (1/2)J_l w^2 to (1/2)M v^2 gives J_l = M(v/w)^2 = 0.422 kg-m^2. (Key point: Referred inertia)

3. What inertia is used to find the accelerating torque?
   A. Load inertia only (0.422)
   B. Motor inertia only (0.1)
   C. Motor plus referred load inertia (0.522 kg-m^2)
   D. Difference of the two
   Answer: C. The motor must accelerate both its own inertia and the referred load inertia: 0.1 + 0.422 = 0.522. (Key point: Total inertia)

4. What torque accelerates the belt in 8 s (friction neglected)?
   A. About 8.2 N-m, from J x (125.66/8)
   B. About 82 N-m
   C. About 0.52 N-m
   D. About 15.7 N-m
   Answer: A. T = J d(omega)/dt = 0.522 x 15.71 = 8.20 N-m. (Key point: Accelerating torque)

## Related in Utilization of Electrical Energy

- [Belt conveyor (2.2 m/s) load inertia and torque](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Abelt-conveyor-2-2ms-inertia-torque) — Utilization of Electrical Energy, chapter 2
- [Winch lifting 500 kg: motor torque and inertia](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Awinch-500kg-motor-torque-inertia) — Utilization of Electrical Energy, chapter 2

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