---
title: "Rolling cylinder attached to a spring"
subject: "Engineering Mechanics"
chapter: 7
key: "semester_3/engineering_mechanics:rolling-cylinder-spring-energy"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/rolling-cylinder-spring-energy
site: BCT Engineering
---

# Rolling cylinder attached to a spring

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/rolling-cylinder-spring-energy
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-7

## Question

A solid cylinder of mass 15 kg and radius 0.25 m rolls without slipping on a horizontal surface. Its axle is attached to a spring of stiffness 400 N/m, whose other end is fixed, with the spring along the line of rolling. The cylinder is displaced 0.3 m from the position where the spring is unstretched and released from rest. Using the work-energy method find the speed of the centre and the angular velocity of the cylinder when it passes through the unstretched position.

## Short answer

Using work-energy: initial spring energy ½ k x² = 18 J; final kinetic energy ½ m v² + ½ I_G ω² = ¾ m v² (since ω = v/r). Solution: v = 1.26 m/s, ω = 5.06 rad/s.

## Answer

**Given:** $m = 15$ kg, $r = 0.25$ m, $k = 400$ N/m, initial spring extension $x = 0.3$ m (released from rest). Rolling without slipping, so $\omega = v/r$.

```text
  |/\/\/\/\/\---(o)-->       cylinder rolls; spring is
  |  k = 400                 unstretched when x = 0
```

#### Moment of inertia

$$
I_G = \tfrac12 mr^2 = \tfrac12(15)(0.25)^2 = 0.4688\ \text{kg·m}^2
$$

#### Work done
Friction at the rolling contact, the normal reaction and the weight do no work (the contact point is instantaneously at rest and the motion is horizontal). Only the spring does work, as its extension goes from 0.3 m to zero:

$$
U_{1\to2} = \tfrac12 k x^2 = \tfrac12(400)(0.3)^2 = 18.0\ \text{J}
$$

#### Kinetic energy
Position 1: at rest, $T_1 = 0$.

Position 2 (spring unstretched):

$$
T_2 = \tfrac12 m v^2 + \tfrac12 I_G\omega^2 = \tfrac12 mv^2 + \tfrac12\left(\tfrac12 mr^2\right)\frac{v^2}{r^2} = \tfrac34 mv^2
$$

#### Work-energy equation

$$
0 + 18.0 = \tfrac34(15)v^2 = 11.25\,v^2
$$

$$
v = \sqrt{\frac{18.0}{11.25}} = 1.265\ \text{m/s}
\qquad
\omega = \frac{v}{r} = \frac{1.265}{0.25} = 5.06\ \text{rad/s}
$$

Check of the energy split: translational $\tfrac12mv^2 = 12.0$ J and rotational $\tfrac14 mv^2 = 6.0$ J, total $18.0$ J.

**Answer: v = 1.26 m/s; ω = 5.06 rad/s.**

## Check yourself

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

1. Which force does work on the rolling cylinder?
   A. Friction at the contact
   B. Normal reaction
   C. Weight
   D. The spring
   Answer: D. Friction at a rolling contact, the normal reaction and the weight do no work; only the spring does. (Key point: Forces doing work)

2. What is the work done by the spring moving from 0.3 m extension to zero?
   A. (1/2)(400)(0.3)^2 = 18 J
   B. 400(0.3) = 120 J
   C. (1/2)(400)(0.3) = 60 J
   D. (400)(0.3)^2 = 36 J
   Answer: A. U = (1/2) k x^2 = 18 J. (Key point: Spring work)

3. For a solid cylinder rolling without slipping, the kinetic energy is:
   A. (1/2) m v^2
   B. (3/4) m v^2
   C. m v^2
   D. (1/4) m v^2
   Answer: B. (1/2)mv^2 + (1/2)(1/2 m r^2)(v/r)^2 = (3/4)mv^2. (Key point: Kinetic energy of rolling disc)

4. What are the speed of the centre and angular velocity at the unstretched position?
   A. v = 1.55 m/s, omega = 6.2 rad/s
   B. v = 1.265 m/s, omega = 5.06 rad/s
   C. v = 1.265 m/s, omega = 1.265 rad/s
   D. v = 2.0 m/s, omega = 8 rad/s
   Answer: B. 18 = 11.25 v^2 gives v = 1.265 m/s, and omega = v/r = 5.06 rad/s. (Key point: Result)

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