---
title: "Rod released from horizontal position"
subject: "Engineering Mechanics"
chapter: 6
key: "semester_3/engineering_mechanics:rod-release-from-horizontal-reaction"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/rod-release-from-horizontal-reaction
site: BCT Engineering
---

# Rod released from horizontal position

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

- Asked in: year not given, 6 marks
- Video: [rotating rod released from horizontal](https://www.youtube.com/watch?v=DzofJqIGcaY) — Dynamics Talklets
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/rod-release-from-horizontal-reaction
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-6

## Question

A uniform slender rod AB of mass 6 kg and length 1.5 m is hinged at A to a fixed support. It is held horizontal and released from rest. Find, at the instant of release, the angular acceleration of the rod, the acceleration of its mass centre and the reaction at the hinge.

## Short answer

Moment of inertia about hinge I_A = mL²/3 = 4.50 kg·m². Angular acceleration α = (mg L/2)/I_A = 9.81 rad/s², center acceleration a_G = 7.36 m/s² downward. Hinge reaction: horizontal zero, vertical 14.7 N upward.

## Answer

**Given:** $m = 6$ kg, $L = 1.5$ m, hinge at A, released from rest in the horizontal position.

```text
   A (hinge) o==========o B
             |-- G --|
   W = mg  v  (at L/2)
```

At the instant of release the rod is at rest, so $\omega = 0$ and the centripetal term $\omega^2 (L/2)$ vanishes. The mass centre G has only a tangential (vertical) acceleration.

#### Moment of inertia about A

$$
I_A = \frac{mL^2}{3} = \frac{6(1.5)^2}{3} = 4.50\ \text{kg·m}^2
$$

#### Angular acceleration
Moments about the hinge (only the weight has a moment; the hinge reaction passes through A):

$$
mg\,\frac{L}{2} = I_A\alpha \;\Rightarrow\; \alpha = \frac{44.14}{4.50} = 9.81\ \text{rad/s}^2\ \left(=\frac{3g}{2L}\right)
$$

#### Acceleration of G

$$
a_G = \alpha\,\frac{L}{2} = 9.81 \times 0.75 = 7.357\ \text{m/s}^2\ \text{(downward)}
$$

#### Hinge reaction
Horizontal: $A_x = 0$ (there is no horizontal acceleration or applied horizontal force).

Vertical (downward positive): $mg - A_y = m\,a_G$

$$
A_y = 6(9.81 - 7.357) = 14.72\ \text{N (upward)}
$$

This is one quarter of the weight ($mg/4 = 58.86/4 = 14.72$ N), a known result for a uniform rod released from the horizontal position.

**Answer: α = 9.81 rad/s²; a_G = 7.36 m/s² downward; hinge reaction = 14.7 N upward (A_x = 0).**

## Check yourself

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

1. Why are moments taken about the hinge A to find alpha?
   A. The hinge reaction passes through A so only the weight has a moment
   B. The rod's acceleration is zero at A
   C. The weight passes through A
   D. Because I_A is smaller than I_G
   Answer: A. mg(L/2) = I_A alpha with the unknown reaction eliminated. (Key point: Moment about the hinge)

2. What is the angular acceleration at release?
   A. g/L
   B. 3g/(2L) = 9.81 rad/s^2
   C. 3g/L
   D. g/(2L)
   Answer: B. alpha = mg(L/2)/(mL^2/3) = 3g/(2L) = 9.81 rad/s^2. (Key point: Angular acceleration)

3. Why does G have only a vertical acceleration at the instant of release?
   A. The hinge is smooth
   B. omega = 0, so the omega^2 (L/2) term vanishes
   C. Gravity cancels the centripetal term
   D. The rod is horizontal so alpha = 0
   Answer: B. Released from rest, there is no normal component; a_G = alpha L/2 = 7.36 m/s^2 downward. (Key point: Centripetal term)

4. What is the vertical hinge reaction at release?
   A. mg = 58.9 N
   B. mg/2 = 29.4 N
   C. mg/4 = 14.7 N upward
   D. 0
   Answer: C. A_y = m(g - a_G) = 6(9.81 - 7.357) = 14.7 N, one quarter of the weight. (Key point: Hinge reaction)

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## Related in other subjects

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