Ball strikes suspended rod, constrained motion

Constrained plane motion is plane motion of a rigid body in which the body is restricted by pins, guides, links or contact surfaces, so it cannot move freely. Because of these constraints, the acceleration of the mass centre $\bar a$ and the angular acceleration $\alpha$ are not independent: they are related by kinematic (geometric) equations. These relations, together with $\sum F = m\bar a$ and $\sum M_G = \bar I\alpha$, let us find the unknown accelerations and reactions.

Examples

• Rotation about a fixed pin not through $G$ (non-centroidal rotation), e.g. a rod or compound pendulum swinging about a pin: $a_t = r\alpha$, $a_n = r\omega^2$.
• A wheel or disc rolling without slipping: $\bar v = \omega r$ and $\bar a = \alpha r$.
• A connecting rod in a slider-crank mechanism, whose ends are guided by the crank pin and the piston.
• A ladder or rod sliding with its ends on a wall and the floor, or a collar sliding on a fixed rod.

Impact problem

The rod hangs from the pin $A$ and the ball strikes its centre $G$, $d = 0.5$ m below $A$. During the impact the weights are non-impulsive and the pin reaction passes through $A$, so neither gives a moment about $A$ and the angular momentum about $A$ is conserved.

Data

$$m_r = \frac{50}{9.81} = 5.097\ \text{kg}, \qquad m_b = \frac{10}{9.81} = 1.019\ \text{kg}, \qquad I_A = \frac13 m_rL^2 = \frac13(5.097)(1)^2 = 1.699\ \text{kg m}^2$$

Let $v_b'$ be the ball's velocity after impact (positive in its original direction) and $\omega$ the angular velocity of the rod.

Conservation of angular momentum about A

$$m_b v\,d = I_A\omega + m_b v_b'\,d \quad (1)$$ $$1.019(9)(0.5) = 1.699\,\omega + 1.019\,v_b'(0.5)$$

Coefficient of restitution

The velocity of the rod at the point of impact is $\omega d$:

$$e = \frac{\omega d - v_b'}{v - 0} \;\Rightarrow\; 0.5\omega - v_b' = 0.4(9) = 3.6 \quad (2)$$

Solving

Substituting $v_b' = 0.5\omega - 3.6$ in (1):

$$4.587 = 1.699\omega + 0.5097(0.5\omega - 3.6) \;\Rightarrow\; \omega = \frac{4.587 + 0.5097(3.6)}{1.699 + 0.5097(0.5)} = 3.29\ \text{rad/s}$$

Then $v_b' = 0.5(3.29) - 3.6 = -1.96$ m/s.

Answer: the angular velocity of the rod just after impact is $\omega = 3.29$ rad/s; the ball rebounds at $1.96$ m/s, opposite to its original direction ($v_b' = -1.96$ m/s).