---
title: "0.45 kg collar and spring on circular rod"
subject: "Applied Mechanics (Dynamics)"
chapter: 2
key: "semester_3/applied_mechanics_dynamics:collar-spring-circular-rod-0-45kg"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/collar-spring-circular-rod-0-45kg
site: BCT Engineering
---

# 0.45 kg collar and spring on circular rod

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

- Asked in: 2078 Bhadra, 8 marks
- Video: [Example 13-4 - Collar and Spring Dynamics | Chapter: 13 | Kinetics of a Particle](https://www.youtube.com/watch?v=GCABbbGOMbc) — XcelTube
- 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-circular-rod-0-45kg
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2

## Question

A 0.45 kg collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 127 mm and a constant $K = 146$ N/m. Knowing that the collar is released from being held at A, determine the speed of the collar and the normal force between the collar and the rod as the collar passes through B. [Figure: circular rod of radius $r$ with centre O; A at the right end of the horizontal diameter, B at the bottom; spring from a wall point C, 178 mm to the left of the circle, to the collar; 127 mm marked on the radius.]

## Short answer

Using conservation of energy between A (release) and B (lowest point), the speed at B is found to be 4.39 m/s. At B, summing radial forces gives the normal reaction between collar and rod as 61.2 N directed toward the centre.

## Answer

**Geometry (read from the figure):** radius $r=0.127$ m, centre O. Collar released at A (right end of the horizontal diameter), B is the lowest point. The spring is fixed at C, $0.178$ m to the left of the circle, on the horizontal line through O. Take O as origin: A $=(0.127,0)$, B $=(0,-0.127)$, C $=(-0.305,0)$. Undeformed length $0.127$ m, $k=146$ N/m, $m=0.45$ kg.

#### Spring lengths
$$
\begin{aligned}
L_A&=0.127+0.305=0.432\ \text{m}\ \Rightarrow\ x_A=0.432-0.127=0.305\ \text{m}\\
L_B&=\sqrt{0.305^{2}+0.127^{2}}=0.3304\ \text{m}\ \Rightarrow\ x_B=0.2034\ \text{m}
\end{aligned}
$$

#### Speed at B (conservation of energy, datum at B)
$$
mg\,r+\tfrac12kx_A^{2}=\tfrac12mv_B^{2}+\tfrac12kx_B^{2}
$$

$$
0.45(9.81)(0.127)+\tfrac12(146)(0.305)^{2}=\tfrac12(0.45)v_B^{2}+\tfrac12(146)(0.2034)^{2}
$$

$$
0.5607+6.7908=0.225v_B^{2}+3.0200\ \Rightarrow\ v_B^{2}=19.25,\ \ v_B=4.388\ \text{m/s}
$$

#### Normal force at B
Spring force $F_s=kx_B=146(0.2034)=29.69$ N, directed from B towards C. Its vertical (upward) component is

$$
F_{s,y}=29.69\cdot\frac{0.127}{0.3304}=11.41\ \text{N}
$$

The horizontal component is tangential, so it does not enter the normal equation. Newton's second law toward the centre O (upward) at B:

$$
N+F_{s,y}-mg=\frac{mv_B^{2}}{r}
$$

$$
N=\frac{0.45(19.25)}{0.127}+0.45(9.81)-11.41=68.2+4.41-11.41=61.2\ \text{N}
$$

**Answer: $v_B\approx4.39$ m/s; normal force $N\approx61.2$ N acting upward, towards O.**

## Check yourself

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

1. What is the spring's deformation at A?
   A. 0.127 m
   B. 0.432 m
   C. 0.305 m
   D. 0.178 m
   Answer: C. L_A = 0.127 + 0.305 = 0.432 m, so x_A = 0.432 - 0.127 = 0.305 m. (Key point: Spring length at A)

2. Which statement describes the energy balance from A to B (datum at B)?
   A. Gravity PE lost at A plus spring energy at A equals KE plus spring energy at B
   B. KE at B equals only the gravity PE lost
   C. Spring energy at B is zero since the rod is smooth
   D. Energy is lost because the normal force acts
   Answer: A. mgr + (1/2)k x_A^2 = (1/2)m v_B^2 + (1/2)k x_B^2, giving v_B about 4.39 m/s. (Key point: Energy equation)

3. Which component of the spring force enters the normal-direction equation at B?
   A. The horizontal component, 29.69 cos(angle)
   B. The full spring force 29.69 N
   C. None, since the spring is stretched
   D. The vertical component, F_s times 0.127/0.3304 = 11.41 N
   Answer: D. At B the normal direction is vertical, so only the vertical component of the spring force counts; the horizontal part is tangential. (Key point: Spring force component)

4. Which equation gives N at B (positive toward O, upward)?
   A. N - F_sy - mg = m v^2/r
   B. N + F_sy - mg = m v^2/r
   C. N + F_sy + mg = m v^2/r
   D. N - mg = m v^2/r
   Answer: B. Toward the centre: N plus the upward spring component minus weight equals m v^2/r, giving N about 61.2 N. (Key point: Normal force at B)

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