---
title: "Winch lifting 500 kg: motor torque and inertia"
subject: "Utilization of Electrical Energy"
chapter: 2
key: "semester_7/utilization_of_electrical_energy:winch-500kg-motor-torque-inertia"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/winch-500kg-motor-torque-inertia
site: BCT Engineering
---

# Winch lifting 500 kg: motor torque and inertia

Utilization of Electrical Energy · Chapter 2: Electric Drive System

- Asked in: 2070 Asar
- 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/winch-500kg-motor-torque-inertia
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-2

## Question

A weight of 500 kg is being lifted up at a uniform speed of 1.5 m/s by a winch which is driven with the help of motor running at a speed of 1000 rpm. The moment of inertia of the motor and winch are 0.5 and 0.3 kg-m² respectively. Calculate the motor torque and the equivalent moment of inertia referred to the motor shaft. In the absence of weight, motor develops a torque of 100 N-m when running at 1000 rpm.

## Short answer

Lifting torque T_w = mgv/ω_m ≈ 70.26 N·m; motor torque T = no-load torque + T_w = 100 + 70.26 ≈ 170.3 N·m. Equivalent inertia J_eq = J_m + J_winch + m(v/ω_m)² = 0.5+0.3+0.1026 =0.9026 ≈0.903 kg·m².

## Answer

**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.

## Check yourself

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

1. What is wm at 1000 rpm?
   A. 16.7 rad/s
   B. 104.72 rad/s
   C. 1000 rad/s
   D. 62.8 rad/s
   Answer: B. wm = 2 pi (1000)/60 = 104.72 rad/s. (Key point: Speed in rad/s)

2. What torque lifts the weight at 1.5 m/s?
   A. 70.26 N-m
   B. 100 N-m
   C. 170.26 N-m
   D. 7357 N-m
   Answer: A. Tw = mgv/wm = 7357.5/104.72 = 70.26 N-m. (Key point: Torque for weight)

3. Motor torque at uniform speed is:
   A. 70.26 N-m
   B. 100 N-m
   C. 29.7 N-m
   D. 170.26 N-m
   Answer: D. No-load 100 N-m plus 70.26 N-m for the weight, no accelerating torque. (Key point: Total motor torque)

4. How is the moving weight's inertia referred to the motor shaft?
   A. m v
   B. m (v/wm)^2, giving 0.1026 kg-m^2
   C. m (wm/v)^2
   D. m g v
   Answer: B. Equal kinetic energy gives J = m (v/wm)^2; total J = 0.9026 kg-m^2. (Key point: Equivalent inertia)

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