--- title: "Impulsive motion, eccentric impact; rod with spring" subject: "Applied Mechanics (Dynamics)" chapter: 6 key: "semester_3/applied_mechanics_dynamics:rod-spring-4kg-impulsive-eccentric" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/rod-spring-4kg-impulsive-eccentric site: BCT Engineering --- # Impulsive motion, eccentric impact; rod with spring Applied Mechanics (Dynamics) · Chapter 6: Plane motion of rigid bodies: energy and momentum methods - Asked in: 2074 Chaitra, 2+6 marks - Video: [Collision of vertical moving rod with frictionless ground(Angular Impulse equation) Good level](https://www.youtube.com/watch?v=EPFQY30Dxzc) — APNA PHYSICS IIT JEE By RKH Sir BTECH IIT D - 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-impulsive-eccentric - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6 ## Question Define impulsive motion and eccentric impact. 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 of total length 600 mm with pivot B 120 mm above C; spring from C to a fixed support D, 350 mm horizontally from B.] ## Short answer Impulsive motion results from large forces over very short time, with positions unchanged. Eccentric impact occurs when the line of impact misses the mass centre, causing rotation. For the rod, work-energy yields angular velocity 11.13 rad/s after 90° (spring and gravity work sum to 15.46 J, I_B = 0.2496 kg·m^2). ## Answer **Impulsive motion** is motion produced by very large forces acting for a very short time (such as a hammer blow or a bullet strike), so that a finite change of velocity occurs in a negligible time. The impulse $\int F\,dt$ of such a force is finite. Non-impulsive forces such as weights and spring forces are neglected during the impact, and the position of the body is taken as unchanged. **Eccentric impact** occurs when the line of impact (the common normal at the point of contact) does not pass through the mass centre of one or both bodies, so the bodies rotate as well as translate after impact. The coefficient of restitution is applied to the relative velocities of the contact points along the line of impact, and the angular momentum principle is used about the pivot or the mass centre. #### Numerical part #### Assumed geometry Rod $AC$ is 0.600 m long: $A$ is 0.480 m above the pivot $B$ and $C$ is 0.120 m below it. The fixed support $D$ is level with $B$, 0.350 m horizontally from $B$. The rod turns through $90^\circ$ until it is horizontal, with $C$ swinging 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 $B$ at the start, and level with $B$ (0.180 m to the side of 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 (the rod turns so that $C$ swings towards the spring support $D$; clockwise as drawn, with $D$ on the left).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What characterises impulsive motion? A. Very large forces act for a very short time, giving a finite velocity change B. A small force acts for a long time C. A constant force acts on a body at rest D. Motion with constant acceleration Answer: A. Impulsive forces are very large and short-lived, so the impulse integral F dt is finite and the velocity changes in a negligible time. (Key point: Impulsive motion) 2. During an impact, which forces are neglected and what happens to the body's position? A. The impact force is neglected B. Weights and spring forces are neglected, and the position does not change during the impact C. Only friction is neglected D. No force is neglected, and the position changes Answer: B. Non-impulsive forces such as weight and spring force are negligible compared with the impulsive force, and the body does not move during the very short impact. (Key point: Non-impulsive forces during impact) 3. When is an impact called eccentric? A. The bodies have equal masses B. The collision is perfectly elastic C. The line of impact does not pass through the mass centre of at least one body, so rotation occurs D. The bodies stick together Answer: C. If the common normal misses the mass centre, the body rotates as well as translates after the impact. (Key point: Eccentric impact) 4. In the numerical part, what gives (1/2) I_B w^2 = 15.46 J? A. Spring energy released 8.40 J plus gravity work 7.06 J B. Spring energy 9.68 J only C. Gravity work 7.06 J only D. Spring energy 8.40 J minus gravity work 7.06 J Answer: A. The spring goes from 9.68 J to 1.28 J (8.40 J released) and G falls 0.18 m (7.06 J), and both add to the kinetic energy, so w = 11.13 rad/s. (Key point: Spring work in the rod problem) ## Close variants set in other exams - [Rod with spring released, 90 degree rotation](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Arod-spring-4kg-pivot-b) — Applied Mechanics (Dynamics), chapter 6, 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 - [Rod and spring, angular velocity at vertical](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Arod-spring-pivot-15kg-explain-impulse) — Applied Mechanics (Dynamics), chapter 6 - [Rod pivoted at O, spring compressed 25 mm](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Arod-spring-pivot-15kg-1-5m) — Applied Mechanics (Dynamics), chapter 6 - [Two slender rods released, angular velocity of AB](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Atwo-rods-released-beta-50) — Applied Mechanics (Dynamics), chapter 6 - [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 - [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 other subjects - [Rod released from horizontal position](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arod-release-from-horizontal-reaction) — Engineering Mechanics, chapter 6 - [Rolling cylinder attached to a spring](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Arolling-cylinder-spring-energy) — Engineering Mechanics, chapter 7