---
title: "Rod with spring released, 90 degree rotation"
subject: "Applied Mechanics (Dynamics)"
chapter: 6
key: "semester_3/applied_mechanics_dynamics:rod-spring-4kg-pivot-b"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/rod-spring-4kg-pivot-b
site: BCT Engineering
---

# Rod with spring released, 90 degree rotation

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

- Asked in: 2078 Bhadra, 6 marks
- Video: [Work and Potential Energy Dynamics Example](https://www.youtube.com/watch?v=FEJqlnlfink) — STEM Course Prep
- 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-4kg-pivot-b
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6

## Question

A slender 4 kg rod can rotate in a vertical plane about a pivot at B. A spring of constant $k = 400$ N/m and of unstretched length 150 mm is attached to the rod as shown. Knowing that the rod is released from rest in the position shown, determine its angular velocity after it has rotated through $90^\circ$. [Figure: vertical rod AC, 480 mm above and 120 mm below the pivot B; spring from C to a fixed support D, 350 mm horizontally from B.]

## Short answer

Applying work-energy, initial spring extension energy 9.68 J, final 1.28 J, and gravity work 7.06 J combine to 15.46 J of rotational KE. With moment of inertia about B 0.2496 kg·m^2, angular velocity after 90° rotation is 11.13 rad/s.

## Answer

#### Assumed geometry
Rod $AC$ is 0.600 m long: $A$ is 0.480 m above the pivot $B$, $C$ is 0.120 m below it. The fixed support $D$ is at the level of $B$, 0.350 m horizontally from $B$. The rod turns through $90^\circ$ so that it becomes horizontal and $C$ moves towards $D$.

```text
            A
            |   0.48 m
            |
   D .......B (pivot)
     <0.35>  |   0.12 m
            C
```

Work-energy principle: $T_1 + V_1 = T_2 + V_2$ with $T_1 = 0$.

#### Spring
$$
\begin{aligned}
\text{Position 1: } CD_1 &= \sqrt{0.350^2 + 0.120^2} = 0.3700\ \text{m}, & x_1 &= 0.3700 - 0.150 = 0.2200\ \text{m}\\
\text{Position 2: } CD_2 &= 0.350 - 0.120 = 0.2300\ \text{m}, & x_2 &= 0.2300 - 0.150 = 0.0800\ \text{m}
\end{aligned}
$$

$$
V_{e1} = \tfrac12(400)(0.2200)^2 = 9.68\ \text{J}, \qquad V_{e2} = \tfrac12(400)(0.0800)^2 = 1.28\ \text{J}
$$

#### Gravity
The centre $G$ of the rod is $0.300$ m from $A$, i.e. $0.480 - 0.300 = 0.180$ m above $B$ initially, and $0.180$ m beside $B$ (level with it) at the end. So $G$ falls $0.180$ m:

$$
W_g = mgh = 4(9.81)(0.180) = 7.06\ \text{J}
$$

#### Moment of inertia about B
$$
I_B = \frac{1}{12}(4)(0.6)^2 + 4(0.18)^2 = 0.2496\ \text{kg m}^2
$$

#### Energy equation
$$
0 + V_{e1} + V_{g1} = \tfrac12 I_B\omega^2 + V_{e2} + V_{g2}, \quad V_{g1}-V_{g2} = W_g \;\Rightarrow\; \tfrac12I_B\omega^2 = 9.68 - 1.28 + 7.06 = 15.46\ \text{J}
$$

$$
\omega = \sqrt{\frac{2(15.46)}{0.2496}} = 11.13\ \text{rad/s}
$$

**Answer: $\omega \approx 11.13$ rad/s (counter-clockwise, towards the spring support).**

## Check yourself

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

1. How is the spring deformation x found at each position?
   A. x = distance from C to D minus the unstretched length 150 mm
   B. x = distance from C to D
   C. x = 350 mm - 150 mm in both positions
   D. x = rod length - 150 mm
   Answer: A. The deformation is the current spring length minus the natural length: 0.370 - 0.150 = 0.220 m initially and 0.230 - 0.150 = 0.080 m finally. (Key point: Spring extension)

2. What is the change in spring energy from position 1 to position 2?
   A. Increases by 8.40 J
   B. It falls from 9.68 J to 1.28 J, releasing 8.40 J
   C. It falls from 9.68 J to 0
   D. It stays constant
   Answer: B. V_e = (1/2)(400)x^2 gives 9.68 J and 1.28 J, so 8.40 J is released and added to the kinetic energy. (Key point: Spring energy released)

3. How far does the rod's centre G fall when the rod turns 90 degrees about B, and why?
   A. 0.48 m
   B. 0.30 m
   C. 0.18 m, because G starts 0.18 m above B and ends level with B
   D. 0.12 m
   Answer: C. G is 0.300 m from A, so it is 0.480 - 0.300 = 0.180 m above B at first and level with B at the end, so W_g = 4(9.81)(0.18) = 7.06 J. (Key point: Gravity work)

4. Which combination gives (1/2) I_B w^2, with I_B = 0.2496 kg m^2?
   A. 9.68 - 1.28 - 7.06
   B. 9.68 + 1.28 + 7.06
   C. 9.68 - 1.28 + 7.06 = 15.46 J
   D. 7.06 only
   Answer: C. Both the spring (released energy 8.40 J) and gravity (7.06 J) do positive work, so (1/2) I_B w^2 = 15.46 J and w = 11.13 rad/s. (Key point: Energy equation and inertia)

## Close variants set in other exams

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