2.7 Physics for Engineers: Kinematics, Newton's Laws, Work-Energy, and Momentum

2.7 Physics for Engineers: Kinematics, Newton's Laws, Work-Energy, and Momentum

"Physics for Engineers" is one of the 15 AMSTHC areas in the 2022 TOS. It has five one-item competencies:

1. Apply Newton's laws of motion to real-world problems.
2. Find the work done by variable forces, including Hooke's law.
3. Use conservation of energy.
4. Assess problems in linear and angular momentum.
5. Develop calculus-based solutions in statics and kinematics.

These dynamics skills also underpin hydraulics (momentum of jets), structures (impact and vibration) and transportation (braking and stopping distance).

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Kinematics of Particles

Calculus definitions.

$v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2 s}{dt^2} = v\frac{dv}{ds}$

The last form is useful when acceleration is given as a function of position.

Constant acceleration:

$v = v_0 + at, \qquad s = v_0 t + \tfrac{1}{2}at^2, \qquad v^2 = v_0^2 + 2as$

Example (calculus-based). A particle moves with $s = t^3 - 6t^2 + 9t$ meters.

• Velocity: $v = 3t^2 - 12t + 9 = 3(t-1)(t-3)$, so it stops at $t = 1$ and $t = 3\text{ s}$.
• Acceleration: $a = 6t - 12$, which is zero at $t = 2\text{ s}$.
• Positions: $s(0) = 0$, $s(1) = 4$, $s(3) = 0$, $s(4) = 4$. The total distance traveled in 4 s is $4 + 4 + 4 = 12\text{ m}$, even though the displacement is only $4\text{ m}$.

Projectile motion (no air resistance) splits into constant horizontal velocity and constant vertical acceleration $g$:

$x = v_0\cos\theta\, t, \qquad y = v_0\sin\theta\, t - \tfrac{1}{2}gt^2$

On level ground, the range is $R = v_0^2 \sin 2\theta / g$, a maximum at $\theta = 45^\circ$. The maximum height is $H = (v_0\sin\theta)^2/(2g)$.

Curvilinear motion. Normal (centripetal) acceleration is $a_n = v^2/\rho$, the basis of highway superelevation.

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Newton's Laws and Friction

1. A body remains at rest or in uniform motion unless acted on by a net force.
2. $\sum F = ma$, applied along each axis.
3. Action and reaction are equal and opposite.

D'Alembert's principle treats $-ma$ as an inertia force, turning a dynamics problem into a statics problem.

Block on an incline with angle $\theta$ and kinetic friction coefficient $\mu$:

• Sliding down: $a = g(\sin\theta - \mu\cos\theta)$.
• Sliding up after an initial push: deceleration $= g(\sin\theta + \mu\cos\theta)$.

Example. A crate slides down a $30^\circ$ chute with $\mu = 0.25$. Then $a = 9.81(0.500 - 0.25 \times 0.866) = 2.78\text{ m/s}^2$. Starting from rest, after $5\text{ m}$ its speed is $v = \sqrt{2(2.78)(5)} = 5.27\text{ m/s}$.

Connected bodies. For a mass $m_1$ on a smooth table pulled by a hanging mass $m_2$ over a frictionless pulley, $a = m_2 g/(m_1 + m_2)$ and the cord tension is $T = m_1 a$.

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Work Done by Variable Forces and Hooke's Law

$W = \int_{s_1}^{s_2} F\,ds$

For a linear spring obeying Hooke's law, $F = kx$, the work to stretch it from $x_1$ to $x_2$ is the area under the force-deformation line:

$W = \tfrac{1}{2}k\left(x_2^2 - x_1^2\right)$

Example. A spring with $k = 4{,}000\text{ N/m}$ is stretched from $0.05$ to $0.15\text{ m}$. $W = 0.5(4{,}000)(0.0225 - 0.0025) = 40\text{ J}$. Equivalently, average force times displacement gives $\tfrac{1}{2}[4{,}000(0.05) + 4{,}000(0.15)](0.10) = 400(0.10) = 40\text{ J}$, which is exact for a linear spring.

Gravity. Lifting a weight $W$ a height $h$ requires $W h$, whatever the path.

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Conservation of Energy and Power

When only conservative forces (gravity and springs) do work:

$T_1 + V_1 = T_2 + V_2, \qquad T = \tfrac{1}{2}mv^2, \qquad V_g = mgh, \qquad V_e = \tfrac{1}{2}kx^2$

With friction or other losses, use the work-energy principle: $T_1 + V_1 + U_{1\to2,\text{nonconservative}} = T_2 + V_2$, where friction work is negative.

Example. A $2\text{ kg}$ block falls from rest $0.80\text{ m}$ onto a spring with $k = 2{,}000\text{ N/m}$. Find the maximum compression $x$:

$mg(0.80 + x) = \tfrac{1}{2}kx^2 \implies 19.62(0.80 + x) = 1{,}000x^2$

Solving the quadratic gives $x = 0.135\text{ m}$.

Power is the rate of doing work: $P = dW/dt = Fv$, with $1\text{ hp} = 746\text{ W}$.

Example. A hoist raises $500\text{ kg}$ at a constant $0.5\text{ m/s}$. $P = mgv = 500(9.81)(0.5) = 2.45\text{ kW}$ delivered to the load. At 80% efficiency, the motor needs $3.07\text{ kW}$.

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Linear Impulse and Momentum

$\int F\,dt = m v_2 - m v_1$

Impulse equals the change in momentum. When no external impulse acts on a system, its total momentum is conserved.

Collisions. Momentum is conserved along the line of impact:

$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$

The coefficient of restitution is:

$e = \frac{v_2' - v_1'}{v_1 - v_2}$

It equals 1 for a perfectly elastic impact (kinetic energy conserved) and 0 for a perfectly plastic impact (the bodies move together).

Example. A $1{,}500\text{ kg}$ car at $20\text{ m/s}$ strikes a stationary $1{,}000\text{ kg}$ car, and they lock together ($e = 0$):

• Common velocity: $v' = 1{,}500(20)/2{,}500 = 12\text{ m/s}$.
• Kinetic energy falls from $300\text{ kJ}$ to $\tfrac{1}{2}(2{,}500)(12)^2 = 180\text{ kJ}$.
• So $120\text{ kJ}$ is dissipated in deformation.

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Angular Momentum and Rotation

For a rigid body rotating about a fixed axis:

$\sum M = I\alpha, \qquad T = \tfrac{1}{2}I\omega^2, \qquad H = I\omega$

The mass moment of inertia $I$ is, for example, $\tfrac{1}{2}mr^2$ for a solid disk and $mr^2$ for a thin ring. The angular impulse $\int M\,dt$ equals the change in angular momentum. When no external moment acts, $I_1\omega_1 = I_2\omega_2$.

Rolling without slipping. For a body of radius $r$ rolling down an incline, $v = \omega r$, and the total kinetic energy is $\tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2$. A solid cylinder ($I = \tfrac{1}{2}mr^2$) rolling from rest down a height $h$ reaches:

$v = \sqrt{\tfrac{4}{3}gh}$

That is slower than a frictionless sliding block, which reaches $\sqrt{2gh}$, because part of the energy goes into rotation.