Impulsive motion, eccentric impact; rod with spring

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 $D$.

            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).