--- title: "Rod and spring, angular velocity at vertical" subject: "Applied Mechanics (Dynamics)" chapter: 6 key: "semester_3/applied_mechanics_dynamics:rod-spring-pivot-15kg-explain-impulse" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/rod-spring-pivot-15kg-explain-impulse site: BCT Engineering --- # Rod and spring, angular velocity at vertical Applied Mechanics (Dynamics) · Chapter 6: Plane motion of rigid bodies: energy and momentum methods - Asked in: 2081 Bhadra, 2+6 marks - Video: [Rotational Dynamics, Bullet Collides with Rod, System Rotates](https://www.youtube.com/watch?v=gc3eH4rRW0g) — Greg Clements - 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-pivot-15kg-explain-impulse - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6 ## Question Explain impulse momentum principle for rigid body. A rod AB of length 1.5 m having a mass of 15 kg is pivoted about point O which is 0.3 m from the end B. The other end of the rod is pressed against a spring having a spring constant of 300 kN/m until the spring is compressed about 25 mm. The rod is then released from its initial position of rest in horizontal position. Determine the angular velocity of the rod through the vertical position. [Figure: horizontal rod AB, spring under end A, pivot O at 1.2 m from A and 0.3 m from B.] ## Short answer Impulse-momentum for a rigid body equates initial linear and angular momenta plus impulses of external forces to final momenta. For the rod, using work-energy, initial spring energy 93.75 J minus gravity work 66.22 J gives rotational KE; with pivot moment of inertia 5.850 kg·m^2, angular velocity after release is 3.07 rad/s. ## Answer **Impulse-momentum principle for a rigid body:** the momenta of all particles of the body at time $t_1$, plus the impulses of the external forces acting between $t_1$ and $t_2$, equal the momenta at time $t_2$. The momenta of the particles reduce to a linear momentum vector $m\bar{\vec v}$ at the mass centre and an angular momentum $\bar I\omega$ about it: $$ \begin{aligned} m\bar{\vec v}_1 + \sum\int \vec F\,dt &= m\bar{\vec v}_2\\ \bar I\omega_1 + \sum\int M_G\,dt &= \bar I\omega_2 \end{aligned} $$ The moment equation may also be written about a fixed point $O$ (or about any point, with the appropriate momentum terms): $H_{O1} + \sum\int M_O\,dt = H_{O2}$. In scalar form the principle gives three equations (two for the linear part, one for the rotational part). It is the best method when forces are given as functions of time, or when a time interval (rather than a distance) is involved, as in impact problems. ```text (momenta)1 + (impulses) = (momenta)2 m v1 .--. F dt m v2 .--. ---> (G) I w1 + ------> = ---> (G) I w2 '--' '--' ``` #### Numerical part #### Data and choice of method This part asks for a speed after a change of position, so the work-energy method is used: the rod starts from rest, the spring and gravity do work, and the pivot reaction does none. $L = 1.5$ m, $m = 15$ kg, $k = 300$ kN/m $= 3\times10^5$ N/m, compression $x = 0.025$ m. Pivot $O$ is 0.3 m from $B$, so the centre $G$ of the rod is $$ d = \frac{L}{2} - 0.3 = 0.75 - 0.3 = 0.45\ \text{m from } O \text{ on the } A \text{ side} $$ ```text B ===O=====G========== A 0.3 0.45 ^ spring ``` Moment of inertia about $O$: $$ I_O = \frac{1}{12}mL^2 + md^2 = \frac{1}{12}(15)(1.5)^2 + 15(0.45)^2 = 5.850\ \text{kg m}^2 $$ #### Position 1 (horizontal, at rest) and position 2 (vertical) When the rod is released, the spring pushes end $A$ up, so the rod swings with $A$ moving up and $G$ rises by $d$ by the time the rod is vertical. $T_1 = 0$. - Elastic energy released by the spring: $V_e = \tfrac12 kx^2 = \tfrac12(3\times10^5)(0.025)^2 = 93.75$ J - Gain in potential energy of the rod: $V_g = mgd = 15(9.81)(0.45) = 66.22$ J $$ T_1 + V_1 = T_2 + V_2 \;\Rightarrow\; 0 + V_e = \tfrac12 I_O\omega^2 + V_g $$ $$ \tfrac12 I_O\omega^2 = 93.75 - 66.22 = 27.53\ \text{J} $$ $$ \omega = \sqrt{\frac{2(27.53)}{5.850}} = 3.07\ \text{rad/s} $$ **Answer: $\omega \approx 3.07$ rad/s (counter-clockwise, with $A$ moving upward) as the rod passes the vertical.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The impulse-momentum principle for a rigid body says: A. Momenta at t1 plus external impulses equal momenta at t2 B. Work equals change in kinetic energy C. Momentum is always conserved D. Force times distance equals impulse Answer: A. Initial momenta plus impulses of external forces between t1 and t2 equal final momenta. (Key point: Impulse-momentum statement) 2. Why is the work-energy method used for the rod passing the vertical? A. Time is given B. Spring and gravity do work, the pivot reaction does none, and the rod starts from rest C. The force is a function of time D. It is an impact problem Answer: B. Distance and energy are involved, and the pivot does no work, so T1 + V1 = T2 + V2 applies. (Key point: Method choice) 3. With d = 0.45 m, m = 15 kg, L = 1.5 m, I_O = (1/12) m L^2 + m d^2 is: A. 2.81 kg m^2 B. 3.04 kg m^2 C. 5.850 kg m^2 D. 8.0 kg m^2 Answer: C. 2.8125 + 3.0375 = 5.850 kg m^2. (Key point: Inertia about O) 4. Spring energy is 93.75 J and the rod's potential gain is 66.22 J. Then omega is: A. 3.07 rad/s B. 4.0 rad/s C. 5.66 rad/s D. 1.5 rad/s Answer: A. (1/2) I_O omega^2 = 27.53 J, so omega = sqrt(2 x 27.53 / 5.85) = 3.07 rad/s. 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