---
title: "Rod and spring, angular velocity and reaction"
subject: "Applied Mechanics (Dynamics)"
chapter: 6
key: "semester_3/applied_mechanics_dynamics:rod-spring-pivot-15kg-2-5m-reaction"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/rod-spring-pivot-15kg-2-5m-reaction
site: BCT Engineering
---

# Rod and spring, angular velocity and reaction

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

- Asked in: 2074 Chaitra, 8 marks
- Video: [Dynamics Example F7-12: Rotation about a fixed point for kinetics of a rigid body](https://www.youtube.com/watch?v=lcaxJqvNetk) — Mechanical Engineering with Dr. Sanei
- 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-2-5m-reaction
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6

## Question

A 15 kg slender rod pivots about the point O. The other end is pressed against a spring ($k = 300$ kN/m) until the spring is compressed one inch and the rod is in a horizontal position. If the rod is released from this position, determine its angular velocity and the reaction at the pivot as the rod passes through a vertical position. [Figure: horizontal rod AB of length 2.5 m, pivot O 0.5 m from B, spring under end A.]

## Short answer

Energy method: spring energy 96.8 J plus gravity work 110.4 J equals rotational KE, giving I_O = 16.250 kg·m^2 and angular velocity 5.05 rad/s. At vertical, centripetal acceleration yields pivot reaction of 434 N upward, no horizontal component.

## Answer

#### Data and assumptions
$L = 2.5$ m, $m=15$ kg, $k = 300$ kN/m, compression $x = 1$ in $= 0.0254$ m. Pivot $O$ is 0.5 m from end $B$ (taken as the end resting on the spring), so the centre $G$ is $d = 1.25 - 0.5 = 0.75$ m from $O$ on the free (long) side.

```text
  spring               O                  G
  B ====================o================== A
   |<---- 0.5 m ---->|<------ 2.0 m ------>|
```

The spring pushes the short end up, so the long side swings down and $G$ falls by $d$ on reaching the vertical. (If the spring pressed the long end instead, the spring energy would be only 96.8 J, less than the 110.4 J needed to lift $G$ through 0.75 m, and the rod could not reach the vertical.)

$$
I_O = \frac{1}{12}mL^2 + md^2 = \frac{1}{12}(15)(2.5)^2 + 15(0.75)^2 = 16.250\ \text{kg m}^2
$$

#### Angular velocity (work-energy)
$$
V_e = \tfrac12kx^2 = \tfrac12(3\times10^5)(0.0254)^2 = 96.77\ \text{J}, \qquad W_g = mgd = 15(9.81)(0.75) = 110.36\ \text{J}
$$

$$
T_1 + U_{1\to2} = T_2:\quad 0 + 96.77 + 110.36 = \tfrac12 I_O\omega^2
$$

$$
\omega = \sqrt{\frac{2(207.14)}{16.250}} = 5.05\ \text{rad/s}
$$

#### Reaction at the pivot in the vertical position
At the vertical, $G$ is directly below $O$. The weight passes through $O$, so $\sum M_O = 0$ and $\alpha = 0$: there is no tangential acceleration and no horizontal reaction. $G$ has only the normal (centripetal) acceleration $\bar a_n = \omega^2 d$ directed up towards $O$:

$$
\bar a_n = \omega^2 d = (5.05)^2(0.75) = 19.12\ \text{m/s}^2
$$

$$
\sum F_y = m\bar a_n:\quad R - mg = m\omega^2 d
$$

$$
R = 15(9.81 + 19.12) = 434\ \text{N}
$$

**Answer: $\omega \approx 5.05$ rad/s; reaction at $O$ is vertical, $R \approx 434$ N upward (horizontal component zero).**

## Check yourself

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

1. In this problem the spring pushes the short end up. What happens to the centre G on reaching the vertical, and how does that enter the energy equation?
   A. G rises by 0.75 m, so gravity work is subtracted
   B. G falls by 0.75 m, so gravity does positive work added to the spring energy
   C. G stays at the same height, so gravity does no work
   D. G falls by 1.25 m
   Answer: B. The long side swings down, so G drops d = 0.75 m and gravity adds mgd = 110.36 J to the 96.77 J of spring energy. (Key point: Direction of G's motion)

2. Which equation gives the angular velocity at the vertical?
   A. 96.77 + 110.36 = (1/2) I_O w^2
   B. 96.77 - 110.36 = (1/2) I_O w^2
   C. 96.77 = (1/2) I_O w^2 + 110.36
   D. 110.36 = (1/2) I_O w^2
   Answer: A. Both the spring and gravity do positive work starting from rest, so the total 207.14 J becomes kinetic energy, giving w = 5.05 rad/s. (Key point: Work-energy equation)

3. At the vertical position, why is there no horizontal reaction at O?
   A. The spring balances it
   B. The weight passes through O, so the moment about O is zero, alpha = 0 and G has no tangential acceleration
   C. The rod has stopped momentarily
   D. The angular velocity is zero
   Answer: B. With G directly below O, gravity gives no moment about O, so alpha = 0 and the only acceleration of G is the normal (centripetal) one, which is vertical. (Key point: Pivot reaction at vertical)

4. Which equation gives the vertical pivot reaction R when G is directly below O?
   A. R = mg
   B. R + mg = m w^2 d
   C. R - mg = m w^2 d
   D. R = m w^2 d only
   Answer: C. G accelerates up towards O with a_n = w^2 d, so R - mg = m w^2 d, giving R = 15(9.81 + 19.12) = 434 N. (Key point: Reaction magnitude)

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