9.2 Newton's Second Law for Particles and Rigid Bodies

9.2 Newton's Second Law for Particles and Rigid Bodies

Core Dynamic Governing Principles: Kinetics relates the forces and moments acting on a physical system to the resulting translational and rotational accelerations. The three foundational methods for solving dynamic FE exam problems are Newton's Second Law ($F = ma$), Work-Energy ($T_1 + U_{1-2} = T_2$), and Impulse-Momentum ($m v_1 + \int F dt = m v_2$).

Newton's Second Law for Particles and Planar Rigid Bodies

Newton's Second Law states that the resultant force acting on a particle equals the time rate of change of its linear momentum: $\mathbf{F} = m \mathbf{a}$.

Planar Rigid Body Equations of Motion

For a rigid body of mass $m$ undergoing planar kinetic motion with center of mass $G$ and mass moment of inertia $I_G$:

$\sum F_x = m a_{Gx}, \quad \sum F_y = m a_{Gy}, \quad \sum M_G = I_G \alpha$

If moments are summed about an arbitrary point $P$ other than the mass center $G$, the moment equation becomes:

$\sum M_P = I_G \alpha + m a_{Gx} y_G - m a_{Gy} x_G$

If point $O$ is a fixed axis of rotation, the rotational equation simplifies to:

$\sum M_O = I_O \alpha$

where $I_O = I_G + m d^2$ by the mass parallel axis theorem.

Mass Moment of Inertia Formulas

Mass moment of inertia $I = \int r^2 dm$ measures rotational resistance:

• Slender Rod (length $L$, mass $m$, about center $G$): $I_G = \frac{1}{12} m L^2$; about end $O$: $I_O = \frac{1}{3} m L^2$.
• Thin Circular Disk / Solid Cylinder (radius $R$, mass $m$): $I_G = \frac{1}{2} m R^2$.
• Thin Hoop / Ring (radius $R$, mass $m$): $I_G = m R^2$.
• Sphere (radius $R$, mass $m$): $I_G = \frac{2}{5} m R^2$.

Worked Engineering Problems

Problem 1: Work-Energy Analysis for Rolling Cylinder and Spring Compression

Scenario: A uniform solid cylinder of mass $m = 10.0\text{ kg}$ and radius $R = 0.20\text{ m}$ ($I_G = \frac{1}{2} m R^2$) is released from rest at the top of an incline angled at $\theta = 30^\circ$ to the horizontal. The cylinder rolls down the incline without slipping for a distance $s = 4.0\text{ m}$ before contacting an uncompressed linear spring of stiffness $k = 800\text{ N/m}$. Determine the speed $v_G$ of the cylinder center just before impacting the spring and the maximum compression $\delta$ of the spring when the cylinder temporarily comes to rest.

Solution:

1. State Work-Energy for pure rolling down distance $s = 4.0\text{ m}$: Initial energy $T_1 = 0$. Potential energy drop $\Delta V_g = m g s \sin 30^\circ = (10.0)(9.81)(4.0)(0.5) = 196.2\text{ J}$. Kinetic energy $T_2$ during pure rolling ($v_G = \omega R$): $T_2 = \frac{1}{2} m v_G^2 + \frac{1}{2} I_G \omega^2 = \frac{1}{2} m v_G^2 + \frac{1}{2} \left( \frac{1}{2} m R^2 \right) \left( \frac{v_G}{R} \right)^2 = \frac{1}{2} m v_G^2 + \frac{1}{4} m v_G^2 = \frac{3}{4} m v_G^2$ $T_2 = \frac{3}{4} (10.0) v_G^2 = 7.5 v_G^2$

2. Solve for velocity $v_G$ at spring contact: $7.5 v_G^2 = 196.2 \implies v_G^2 = \frac{196.2}{7.5} = 26.16 \implies v_G = \sqrt{26.16} \approx 5.115\text{ m/s} \approx 5.12\text{ m/s}$

3. Work-Energy from initial rest position to maximum spring compression $\delta$: Total vertical drop $h_{\text{total}} = (s + \delta) \sin 30^\circ = (4.0 + \delta)(0.5)$. $m g (4.0 + \delta)(0.5) = \frac{1}{2} k \delta^2$ $(10.0)(9.81)(0.5)(4.0 + \delta) = \frac{1}{2}(800)\delta^2$ $49.05 (4.0 + \delta) = 400 \delta^2 \implies 196.2 + 49.05 \delta = 400 \delta^2$ $400 \delta^2 - 49.05 \delta - 196.2 = 0$ Using quadratic formula: $\delta = \frac{49.05 + \sqrt{(-49.05)^2 - 4(400)(-196.2)}}{2(400)} = \frac{49.05 + \sqrt{2405.9 + 313,920}}{800} = \frac{49.05 + 562.43}{800} \approx 0.764\text{ m}$

Problem 2: Direct Central Impact and Kinetic Energy Loss Calculation

Scenario: Sphere $A$ ($m_A = 3.0\text{ kg}$) travelling right at $v_{A1} = 8.0\text{ m/s}$ collides head-on with sphere $B$ ($m_B = 5.0\text{ kg}$) travelling left at $v_{B1} = -2.0\text{ m/s}$. The coefficient of restitution is $e = 0.75$. Determine post-impact velocities $v_{A2}$ and $v_{B2}$, and compute total kinetic energy lost during collision $\Delta T$.

Solution:

1. Apply Conservation of Linear Momentum: $m_A v_{A1} + m_B v_{B1} = m_A v_{A2} + m_B v_{B2}$ $(3.0)(8.0) + (5.0)(-2.0) = 3.0 v_{A2} + 5.0 v_{B2} \implies 24.0 - 10.0 = 14.0 = 3.0 v_{A2} + 5.0 v_{B2}$

2. Apply Coefficient of Restitution Equation: $e = \frac{v_{B2} - v_{A2}}{v_{A1} - v_{B1}} \implies 0.75 = \frac{v_{B2} - v_{A2}}{8.0 - (-2.0)} = \frac{v_{B2} - v_{A2}}{10.0}$ $v_{B2} - v_{A2} = 7.50 \implies v_{B2} = v_{A2} + 7.50$

3. Substitute $v_{B2}$ into momentum equation: $3.0 v_{A2} + 5.0(v_{A2} + 7.50) = 14.0 \implies 8.0 v_{A2} + 37.50 = 14.0$ $8.0 v_{A2} = -23.50 \implies v_{A2} = -2.9375\text{ m/s} \approx -2.94\text{ m/s} \quad (\text{rebounds left})$ $v_{B2} = -2.9375 + 7.50 = +4.5625\text{ m/s} \approx +4.56\text{ m/s} \quad (\text{moves right})$

4. Calculate Kinetic Energy Loss $\Delta T$: $T_1 = \frac{1}{2}(3.0)(8.0)^2 + \frac{1}{2}(5.0)(-2.0)^2 = 96.0 + 10.0 = 106.0\text{ J}$ $T_2 = \frac{1}{2}(3.0)(-2.9375)^2 + \frac{1}{2}(5.0)(4.5625)^2 = 12.94 + 52.04 = 64.98\text{ J}$ $\Delta T = T_1 - T_2 = 106.0 - 64.98 = 41.02\text{ J}$ Over 41.0 J of mechanical energy is dissipated as deformation heat and sound.