---
title: "Blocks over a massive pulley"
subject: "Engineering Mechanics"
chapter: 6
key: "semester_3/engineering_mechanics:pulley-blocks-rotational-inertia"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/pulley-blocks-rotational-inertia
site: BCT Engineering
---

# Blocks over a massive pulley

Engineering Mechanics · Chapter 6: Plane Kinetics of Rigid Bodies: Force, Mass and Acceleration

- Asked in: year not given, 8 marks
- Video: [Physics 13.1  Moment of Inertia Application (10 of 11) Acceleration=? When Pulley Has Mass](https://www.youtube.com/watch?v=TiZh9DYZ744) — Michel van Biezen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/pulley-blocks-rotational-inertia
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-6

## Question

Two blocks A (8 kg) and B (4 kg) hang from the ends of a light inextensible rope that passes over a pulley, which is a uniform solid disc of mass 10 kg and radius 0.2 m, mounted on a smooth horizontal axle. The rope does not slip on the pulley. The system is released from rest. Find the acceleration of the blocks, the angular acceleration of the pulley and the tensions in the two parts of the rope.

## Short answer

With a = α r and I = ½ M r², acceleration of blocks a = (m_A - m_B)g/(m_A + m_B + M/2) = 2.31 m/s², angular acceleration α = 11.54 rad/s²; tensions T_A = 60.0 N, T_B = 48.5 N.

## Answer

**Given:** $m_A = 8$ kg, $m_B = 4$ kg, pulley $M = 10$ kg, $r = 0.2$ m. Since $m_A > m_B$, A moves down and B moves up. Pulley moment of inertia:

$$
I = \tfrac12 Mr^2 = \tfrac12(10)(0.2)^2 = 0.20\ \text{kg·m}^2
$$

```text
        ____
       /    \   pulley (M, r)
  T_A |      | T_B
      |      |
     [A]    [B]
   (down)   (up)
```

The rope does not slip, so $a = \alpha r$. The tensions in the two parts differ because the pulley has inertia.

#### Equations
Block A (downward positive): $m_Ag - T_A = m_A a$ (1)

Block B (upward positive): $T_B - m_Bg = m_B a$ (2)

Pulley (moments about the axle): $(T_A - T_B)\,r = I\alpha = \tfrac12 Mr^2\,\dfrac{a}{r}$

$$
T_A - T_B = \tfrac12 Ma = 5a \quad (3)
$$

Adding (1), (2) and (3): $(m_A - m_B)g = (m_A + m_B + \tfrac12 M)\,a$

#### Acceleration

$$
a = \frac{(8 - 4)(9.81)}{8 + 4 + 5} = \frac{39.24}{17} = 2.308\ \text{m/s}^2
$$

#### Angular acceleration

$$
\alpha = \frac{a}{r} = \frac{2.308}{0.2} = 11.54\ \text{rad/s}^2
$$

The pulley turns in the sense that moves A's side down (counter-clockwise in the sketch, where A is on the left).

#### Tensions

$$
T_A = m_A(g - a) = 8(9.81 - 2.308) = 60.01\ \text{N}
$$

$$
T_B = m_B(g + a) = 4(9.81 + 2.308) = 48.47\ \text{N}
$$

Check: $T_A - T_B = 11.54$ N $= 5a$. It agrees.

**Answer: a = 2.31 m/s²; α = 11.54 rad/s²; T_A = 60.0 N; T_B = 48.5 N.**

## Check yourself

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

1. Why are the rope tensions T_A and T_B different here?
   A. Block A is heavier so gravity changes the rope
   B. The pulley has rotational inertia, so a net torque (T_A - T_B) r is needed to accelerate it
   C. The rope stretches
   D. The axle is rough
   Answer: B. With a massless pulley the tensions would be equal; here (T_A - T_B) r = I alpha. (Key point: Unequal tensions)

2. For the disc pulley with no rope slip, what does (T_A - T_B) r = I alpha reduce to?
   A. T_A - T_B = M a
   B. T_A - T_B = (1/2) M a = 5a
   C. T_A - T_B = M g
   D. T_A - T_B = 2 M a
   Answer: B. Using I = (1/2)M r^2 and alpha = a/r gives (1/2)M a. (Key point: Pulley equation)

3. Which expression gives the acceleration?
   A. (m_A - m_B) g / (m_A + m_B) = 3.27 m/s^2
   B. (m_A - m_B) g / (m_A + m_B + M/2) = 2.31 m/s^2
   C. (m_A - m_B) g / (m_A + m_B + M) = 1.78 m/s^2
   D. (m_A + m_B) g / (m_A - m_B + M/2) = 11 m/s^2
   Answer: B. Adding the three equations gives 4(9.81) = 17a, so a = 2.308 m/s^2. (Key point: Acceleration)

4. Which pair of tensions is correct?
   A. T_A = 60.0 N, T_B = 48.5 N
   B. T_A = 48.5 N, T_B = 60.0 N
   C. T_A = T_B = 54.2 N
   D. T_A = 78.5 N, T_B = 39.2 N
   Answer: A. T_A = 8(9.81 - 2.308) = 60.0 N and T_B = 4(9.81 + 2.308) = 48.5 N; the heavier side has the larger tension. (Key point: Tensions)

## Related in Engineering Mechanics

- [Two blocks connected over a pulley](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Ablocks-pulley-friction-acceleration) — Engineering Mechanics, chapter 2

## Related in other subjects

- [Engine test: four-cylinder with rope radius](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aengine-test-rope-radius) — Power Plant Equipment, chapter 5, asked 5×