---
title: "Rod pivoted at O, spring compressed 25 mm"
subject: "Applied Mechanics (Dynamics)"
chapter: 6
key: "semester_3/applied_mechanics_dynamics:rod-spring-pivot-15kg-1-5m"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/rod-spring-pivot-15kg-1-5m
site: BCT Engineering
---

# Rod pivoted at O, spring compressed 25 mm

Applied Mechanics (Dynamics) · Chapter 6: Plane motion of rigid bodies: energy and momentum methods

- Asked in: 2076 Chaitra, 6 marks
- Video: [Dynamics 03_08 Work and Potential Energy Problem with Solution in Kinetics of Particles](https://www.youtube.com/watch?v=Q1iU3jFKJhg) — Kasma Tutorial
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/rod-spring-pivot-15kg-1-5m
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6

## Question

A 15kg slender rod AB is 1.5m long and is pivoted about a point O which is 0.3m from the end B. The other end is pressed against a spring of constant K = 300KN/m until the spring is compressed 25mm. The rod is then in horizontal position. If the rod is released from this position, determine the angular velocity as the rod passed through a vertical position. [Figure: horizontal rod AB of length 1.5 m, pivot O 0.3 m from B, spring under end A.]

## Short answer

Work-energy method: spring compression releases 93.75 J, gravity work is −66.22 J, leaving 27.53 J as rotational KE. With moment of inertia about O 5.850 kg·m^2, angular velocity as the rod passes vertical is 3.07 rad/s.

## Answer

#### Data and choice of method
Energy method (work-energy): the rod starts from rest and the spring and gravity do work; the pivot reaction does none.

$L = 1.5$ m, $m = 15$ kg, $k = 300$ kN/m $= 3\times10^5$ N/m, compression $x = 0.025$ m. Pivot $O$ is 0.3 m from $B$, so the centre $G$ of the rod is

$$
d = \frac{L}{2} - 0.3 = 0.75 - 0.3 = 0.45\ \text{m from } O \text{ on the } A \text{ side}
$$

```text
  B ===O=====G========== A
     0.3   0.45
                       ^ spring
```

Moment of inertia about $O$:

$$
I_O = \frac{1}{12}mL^2 + md^2 = \frac{1}{12}(15)(1.5)^2 + 15(0.45)^2 = 5.850\ \text{kg m}^2
$$

#### Position 1 (horizontal, at rest) and position 2 (vertical)
The spring is released, pushing end $A$ up, so the rod swings with $A$ up and $G$ rises by $d$ in reaching the vertical. $T_1 = 0$.

- Elastic energy released by the spring: $V_e = \tfrac12 kx^2 = \tfrac12(3\times10^5)(0.025)^2 = 93.75$ J
- Gain in potential energy of the rod: $V_g = mgd = 15(9.81)(0.45) = 66.22$ J

$$
T_1 + V_1 = T_2 + V_2 \;\Rightarrow\; 0 + V_e = \tfrac12 I_O\omega^2 + V_g
$$

$$
\tfrac12 I_O\omega^2 = 93.75 - 66.22 = 27.53\ \text{J}
$$

$$
\omega = \sqrt{\frac{2(27.53)}{5.850}} = 3.07\ \text{rad/s}
$$

**Answer: $\omega \approx 3.07$ rad/s (counter-clockwise, with $A$ moving upward) as the rod passes the vertical.**

## Check yourself

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

1. Why is the work-energy method the natural choice for finding the angular velocity at the vertical position?
   A. Spring and gravity do work, the pivot reaction does none, and the speed is wanted at a position, not a time
   B. Angular momentum about O is conserved during the swing
   C. The rod's angular acceleration is constant
   D. The spring force is constant throughout the motion
   Answer: A. The rod starts from rest, only the spring and gravity do work, and the pivot reaction does no work, so T1 + V1 = T2 + V2 gives omega directly. (Key point: Choice of method)

2. The pivot O is 0.3 m from end B of a 1.5 m rod. How far is the centre G from O?
   A. 0.75 m
   B. 0.30 m
   C. 0.45 m
   D. 1.20 m
   Answer: C. G is at L/2 = 0.75 m from B, so it is 0.75 - 0.3 = 0.45 m from O. (Key point: Centre of mass offset)

3. Which expression gives the rod's moment of inertia about the pivot O?
   A. (1/3) m L^2
   B. (1/12) m L^2 + m d^2 with d = 0.45 m
   C. (1/12) m L^2 only
   D. m d^2 only
   Answer: B. The parallel-axis theorem is used: I_O = I_G + m d^2 = 2.8125 + 3.0375 = 5.85 kg m^2. (Key point: Moment of inertia about O)

4. How is the energy balance written between the horizontal and vertical positions?
   A. Spring energy + mgd = (1/2) I_O w^2
   B. Spring energy = (1/2) I_O w^2 + mgd, since G rises by d
   C. (1/2) I_O w^2 = mgd only
   D. Spring energy = (1/2) m w^2 only
   Answer: B. The spring's 93.75 J is partly stored as the rod's potential energy (66.22 J, G rises 0.45 m), and the rest becomes kinetic energy of 27.53 J, giving w = 3.07 rad/s. (Key point: Energy balance)

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