--- title: "Conservation of energy; 5 kg collar on curved rod" subject: "Applied Mechanics (Dynamics)" chapter: 2 key: "semester_3/applied_mechanics_dynamics:collar-spring-vertical-5kg" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/collar-spring-vertical-5kg site: BCT Engineering --- # Conservation of energy; 5 kg collar on curved rod Applied Mechanics (Dynamics) · Chapter 2: Kinetics of particles: Energy and Momentum Methods - Asked in: 2080 Baishakh, 2+6 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-5kg - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2 ## Question Explain about the conservation of energy in study of kinetics of particles with relevant equation. A 5 kg collar is attached to a spring and slides without friction in a vertical plane along a 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 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 Mechanical energy (kinetic + gravitational + elastic) is conserved. With the given spring and rod dimensions, the collar's speed at B is 2.67 m/s, and because C is at the same height and spring extension as A, it stops there (v_C = 0). ## Answer #### Conservation of energy in kinetics of particles When a particle moves under conservative forces only (weight, elastic spring force), the sum of kinetic energy $T=\tfrac12mv^{2}$ and potential energy $V=V_g+V_e$ stays constant: $$ T_1+V_1=T_2+V_2,\qquad V_g=mgy,\quad V_e=\tfrac12kx^{2} $$ where $x$ is the spring deformation from its undeformed length. Forces that do no work (normal reaction of a smooth rod) can be ignored. #### Collar problem **Assumed geometry (figure dimensions 150, 200, 250 mm):** semicircular rod of radius $0.2$ m with A and C at the same level, B $0.2$ m below. The spring is fixed at D on the vertical axis through B, $0.15$ m below the line AC. Then the spring length is $L_A=L_C=\sqrt{0.2^{2}+0.15^{2}}=0.25$ m and $L_B=0.2-0.15=0.05$ m. Undeformed length $0.10$ m; $k=800$ N/m; $m=5$ kg. Spring deformations: $x_A=x_C=0.25-0.10=0.15$ m (stretched) and $x_B=0.05-0.10=-0.05$ m (compressed). #### a) At B Released from rest at A (take B as datum, $h_A=0.2$ m): $$ 0+mgh_A+\tfrac12kx_A^{2}=\tfrac12mv_B^{2}+0+\tfrac12kx_B^{2} $$ $$ 5(9.81)(0.2)+\tfrac12(800)(0.15)^{2}=\tfrac12(5)v_B^{2}+\tfrac12(800)(0.05)^{2} $$ $$ 9.81+9.00=2.5v_B^{2}+1.00\ \Rightarrow\ v_B^{2}=7.124,\ \ v_B=2.669\ \text{m/s} $$ #### b) At C C is at the same height as A and the spring has the same length, so both potential energies are the same as at A: $$ \tfrac12mv_C^{2}=0\ \Rightarrow\ v_C=0 $$ **Answer: $v_B\approx2.67$ m/s and $v_C=0$ for the geometry assumed above. If the figure shows C at a different level or spring length, use the same energy equation with those values.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Under what condition does T1 + V1 = T2 + V2 hold for a particle? A. Only conservative forces (gravity, spring) do work B. Any force acts, as long as the speed is constant C. Friction is present but small D. The particle starts from rest Answer: A. The sum of kinetic and potential energy stays constant only when work is done by conservative forces such as weight and an elastic spring. (Key point: Conservation of energy statement) 2. Why can the normal reaction of the smooth rod be ignored in the energy equation? A. It is always equal to the weight B. It is perpendicular to the motion and does no work C. It is cancelled by the spring force D. It is much smaller than gravity Answer: B. The normal force acts perpendicular to the displacement, so it does no work and does not enter the energy equation. (Key point: Smooth rod normal force) 3. In Ve = (1/2)k x^2, what is x for the collar problem? A. The spring's total length B. The collar's distance from the datum C. The change of length from the undeformed length (0.10 m) D. The distance travelled along the rod Answer: C. x is the deformation from the undeformed length, so x_A = 0.25 - 0.10 = 0.15 m and x_B = 0.05 - 0.10 = -0.05 m. (Key point: Spring deformation) 4. Which energy balance gives v_B for the 5 kg collar (datum at B)? A. 0.5(5)v_B^2 = 5(9.81)(0.2) B. 5(9.81)(0.2) + 0.5(800)(0.15)^2 = 0.5(5)v_B^2 + 0.5(800)(0.05)^2 C. 5(9.81)(0.2) = 0.5(5)v_B^2 + 0.5(800)(0.15)^2 D. 5(9.81)(0.2) + 0.5(800)(0.05)^2 = 0.5(5)v_B^2 + 0.5(800)(0.15)^2 Answer: B. At A the collar has gravity and spring (x = 0.15) energy; at B it has kinetic energy and spring (x = 0.05) energy. This gives v_B about 2.67 m/s. (Key point: Speed at B) ## Close variants set in other exams - [Conservation of energy; 2 kg collar on curved rod](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acollar-spring-vertical-2kg) — 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