---
title: "Conservation of energy; 2 kg collar on curved rod"
subject: "Applied Mechanics (Dynamics)"
chapter: 2
key: "semester_3/applied_mechanics_dynamics:collar-spring-vertical-2kg"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/collar-spring-vertical-2kg
site: BCT Engineering
---

# Conservation of energy; 2 kg collar on curved rod

Applied Mechanics (Dynamics) · Chapter 2: Kinetics of particles: Energy and Momentum Methods

- Asked in: 2074 Asoj, 3+5 marks
- Video: [14-68 Kinetics of Particle: Conservation of Energy Chapter 14: Hibbeler Dynamics | Engineers Academy](https://www.youtube.com/watch?v=SPJZwgyRjbQ) — Engineers Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/collar-spring-vertical-2kg
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2

## Question

a) What is the principle of conservation of energy of a system? Illustrate it with suitable example. b) A 2 kg collar is attached to a spring and slides without friction in a vertical plane along the curved rod ABC. The spring is undeformed when its length is 100 mm and its constant is 800 N/m. If the collar is released at 'A' with no initial velocity, determine its velocity (a) as it passes through 'B' (b) as it reaches 'C'. [Figure: semicircular rod ABC; collar at A at the right end, C at the left end; spring from the collar to the vertical axis through B; dimensions 150 mm, 250 mm and 200 mm.]

## Short answer

Total mechanical energy is conserved when only conservative forces work; for example, a pendulum converts potential energy to kinetic. For the 2 kg collar, energy conservation gives speed 3.45 m/s at B and zero at C, assuming C is at the same level and spring stretch as A.

## Answer

#### a) Principle of conservation of energy
The total mechanical energy of a system, kinetic plus potential, remains constant when only conservative forces (gravity, springs) do work:

$$
T_1+V_1=T_2+V_2
$$

**Example:** a pendulum bob released from rest at height $h$ above its lowest point. At the top it has only potential energy $mgh$; at the bottom only kinetic energy $\tfrac12mv^{2}$. Hence $v=\sqrt{2gh}$. During the swing the energy changes between the two forms but the total is constant (air friction neglected).

#### b) Collar problem
**Assumed geometry (figure dimensions 150, 200, 250 mm):** semicircular rod of radius $0.2$ m, A and C at the same level, B $0.2$ m below them. The spring is fixed at D on the vertical axis through B, $0.15$ m below line AC, so $L_A=L_C=\sqrt{0.2^{2}+0.15^{2}}=0.25$ m and $L_B=0.05$ m. Undeformed length $0.10$ m, $k=800$ N/m, $m=2$ kg.

Deformations: $x_A=x_C=0.15$ m, $x_B=-0.05$ m.

**At B** (datum at B, $h_A=0.2$ m):

$$
2(9.81)(0.2)+\tfrac12(800)(0.15)^{2}=\tfrac12(2)v_B^{2}+\tfrac12(800)(0.05)^{2}
$$

$$
3.924+9.00=v_B^{2}+1.00\ \Rightarrow\ v_B^{2}=11.924,\ \ v_B=3.453\ \text{m/s}
$$

**At C:** same height and same spring deformation as at A, so $\tfrac12mv_C^{2}=0$ and $v_C=0$.

**Answer: $v_B\approx3.45$ m/s; $v_C=0$ for the assumed geometry (use the same energy equation with figure values if C lies at a different level).**

## Check yourself

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

1. In the pendulum example, what is the speed at the lowest point after release from height h?
   A. v = gh
   B. v = sqrt(gh)
   C. v = 2gh
   D. v = sqrt(2gh)
   Answer: D. mgh = (1/2)mv^2 gives v = sqrt(2gh): potential energy converts fully into kinetic energy. (Key point: Pendulum example)

2. What stays constant when only conservative forces do work?
   A. Kinetic energy alone
   B. Potential energy alone
   C. The sum of kinetic and potential energy
   D. Momentum of the particle
   Answer: C. Energy shifts between kinetic and potential forms but T + V stays constant. (Key point: Conservation principle)

3. For the assumed geometry, what is the spring length at B?
   A. 0.25 m
   B. 0.05 m
   C. 0.15 m
   D. 0.35 m
   Answer: B. B is 0.2 m below AC and the spring is fixed 0.15 m below AC, so L_B = 0.2 - 0.15 = 0.05 m (compressed, x = -0.05 m). (Key point: Spring lengths)

4. Why is v_C = 0 for the 2 kg collar?
   A. The rod is rough at C
   B. The spring force stops it
   C. C is at the same height and spring length as A, so no net energy change
   D. The mass is too small
   Answer: C. Gravity and spring energies at C equal those at A, so kinetic energy must be zero again. (Key point: Speed at C)

## Close variants set in other exams

- [Conservation of energy; 5 kg collar on curved rod](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acollar-spring-vertical-5kg) — Applied Mechanics (Dynamics), chapter 2, asked 2×

## Related in Applied Mechanics (Dynamics)

- [0.45 kg collar and spring on circular rod](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acollar-spring-circular-rod-0-45kg) — Applied Mechanics (Dynamics), chapter 2
- [Types of impact; collar on curved rod](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aimpact-types-collar-curved-rod-horizontal) — Applied Mechanics (Dynamics), chapter 2
- [Types of impact; collar on vertical rod](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aimpact-types-collar-spring-vertical) — Applied Mechanics (Dynamics), chapter 2