---
title: "Block lowering a drum, energy method"
subject: "Engineering Mechanics"
chapter: 7
key: "semester_3/engineering_mechanics:drum-block-work-energy-acceleration"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/drum-block-work-energy-acceleration
site: BCT Engineering
---

# Block lowering a drum, energy method

Engineering Mechanics · Chapter 7: Plane Motion of Rigid Bodies: Work and Energy Method

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/drum-block-work-energy-acceleration
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-7

## Question

A drum of mass 20 kg, radius 0.30 m and radius of gyration 0.25 m about its fixed horizontal axis has a light rope wound round it carrying a 10 kg block at the free end. The system is released from rest. Using the work-energy method, find the speed of the block after it has descended 2 m. Hence find the acceleration of the block, the angular acceleration of the drum and the tension in the rope.

## Short answer

Work-energy: 10×9.81×2 = ½ v²(10 + 1.25/0.3²) → v = 4.05 m/s. Acceleration a = v²/(2h) = 4.11 m/s²; angular acceleration α = 13.69 rad/s²; rope tension T = 10(9.81 - 4.11) = 57.0 N.

## Answer

**Given:** drum $M = 20$ kg, $r = 0.30$ m, $k = 0.25$ m; block $m = 10$ kg; fall $h = 2$ m from rest.

```text
      ___
     /   \  drum (fixed axle)
     \___/|
          | rope
         [m]  falls h = 2 m
```

#### Drum moment of inertia

$$
I = Mk^2 = 20(0.25)^2 = 1.25\ \text{kg·m}^2 \qquad \frac{I}{r^2} = 13.889\ \text{kg}
$$

#### Work-energy method
Rope tension does no net work on the system (it acts on both the block and the drum with opposite signs); only the weight of the block does work:

$$
U = mgh = 10(9.81)(2) = 196.2\ \text{J}
$$

Final kinetic energy, with $\omega = v/r$:

$$
T_2 = \tfrac12 mv^2 + \tfrac12 I\omega^2 = \tfrac12 v^2\left(m + \frac{I}{r^2}\right) = \tfrac12 v^2(10 + 13.889)
$$

$$
0 + 196.2 = \tfrac12 v^2(23.889) \;\Rightarrow\; v = 4.053\ \text{m/s}
$$

#### Acceleration of the block
For constant acceleration from rest, $v^2 = 2ah$:

$$
a = \frac{v^2}{2h} = \frac{(4.053)^2}{4} = 4.107\ \text{m/s}^2
$$

Check: $a = \dfrac{mg}{m + I/r^2} = \dfrac{98.1}{23.889} = 4.107$ m/s².

#### Angular acceleration and tension

$$
\alpha = \frac{a}{r} = \frac{4.107}{0.3} = 13.69\ \text{rad/s}^2
\qquad
T = m(g - a) = 10(9.81 - 4.107) = 57.03\ \text{N}
$$

**Answer: v = 4.05 m/s after 2 m; a = 4.11 m/s²; α = 13.69 rad/s²; T = 57.0 N.**

## Check yourself

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

1. Which force does net work on the drum-and-block system?
   A. Rope tension on the block only
   B. Weight of the block only
   C. Tension and weight both
   D. Drum axle reaction
   Answer: B. The tension does equal and opposite work on block and drum, so it cancels; the axle reaction does no work. (Key point: Work in the system)

2. How is the drum moment of inertia obtained?
   A. I = (1/2) M r^2 = 0.9
   B. I = M k^2 = 1.25 kg·m^2
   C. I = M r^2 = 1.8
   D. I = M k = 5
   Answer: B. k is the radius of gyration, so I = 20(0.25)^2 = 1.25 kg·m^2. (Key point: Drum inertia)

3. What is the block speed after descending 2 m?
   A. 6.26 m/s
   B. 4.05 m/s
   C. 3.20 m/s
   D. 8.1 m/s
   Answer: B. 196.2 = (1/2) v^2 (10 + 13.889) gives v = 4.053 m/s. (Key point: Speed of the block)

4. How is the rope tension found after getting a = 4.11 m/s^2?
   A. T = m g = 98.1 N
   B. T = m(g - a) = 57.0 N
   C. T = m(g + a) = 139 N
   D. T = m a = 41.1 N
   Answer: B. For the block, mg - T = ma, so T = 10(9.81 - 4.107) = 57.0 N. (Key point: Tension)

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