Dynamics

This chapter covers the following topics:

• Particle kinematics
• Plane circular motion
• Projectile motion
• Particle kinetics (Newton’s second law)
• Principle of work and energy
• Kinetic and potential energy
• Work, power, and efficiency
• Impulse and momentum
• Angular momentum
• Impact
• Dynamic friction
• Plane motion of a rigid body
• Free and forced vibration
• Torsional vibration

Particle kinematics

Particle kinematics describes motion without considering the forces that cause it. The goal is to describe how a particle’s position, velocity, and acceleration change with time.

You’ll typically express motion using either scalar variables (for straight-line motion) or vectors (for motion in 2D or 3D). Depending on the path, motion may be described in Cartesian, normal-tangential, or polar coordinates. Kinematics gives you the mathematical tools you’ll later use in dynamics.

Rectilinear motion (straight line)

In straight-line motion, position is described by a single coordinate $x(t)$.

• Position: $x(t)$
• Velocity: $v(t)=\frac{dx}{dt}$
• Acceleration: $a(t)=\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$

Constant acceleration equations

These equations apply when acceleration $a$ is constant.

$$v=v_0+at$$

$$x=x_0+v_{0}t+\frac{1}{2}a{t}^2$$

$$v^2=v_0^2+2a(x-x_0)$$

Example: displacement under constant acceleration A car accelerates from rest at $a=2{\text{m/s}}^2$ for $t=8\text{s}$. Find the displacement.

$$x=0+0\cdot 8+\frac{1}{2}(2)(8^2)=64\text{m}$$

Answer: $64\text{m}$

Curvilinear motion

For motion along a curved path, it’s often clearer to work with vectors.

• Position vector: $\vec{r}(t)$
• Velocity: $\vec{v}(t)=\frac{d\vec{r}}{dt}$
• Acceleration: $\vec{a}(t)=\frac{d\vec{v}}{dt}$

Example: velocity and acceleration from a position vector If $\vec{r}(t)=3t\hat{i}+t^2\hat{j}$, differentiate each component with respect to $t$.

$$\vec{v}(t)=3\hat{i}+2t\hat{j}$$

$$\vec{a}(t)=2\hat{j}$$

Answer: $\vec{v}(t)=3\hat{i}+2t\hat{j}$, $\vec{a}(t)=2\hat{j}$

Plane circular motion

Plane circular motion is motion along a circular path in a two-dimensional plane. The acceleration naturally splits into two perpendicular components:

• a tangential component (changes the speed)
• a radial (centripetal) component (changes the direction)

It’s useful to distinguish:

• Uniform circular motion: speed is constant (no tangential acceleration)

• Nonuniform circular motion: speed changes (tangential acceleration is present)

• Tangential velocity:

$$v=r\omega$$

• Tangential acceleration:

  $$a_t=r\alpha$$

• Radial (centripetal) acceleration:

  $$a_r=\frac{v^2}{r}=r{\omega }^2$$

Example: speed and centripetal acceleration of a spinning wheel For a wheel of radius $r=0.5\text{m}$ spinning at $\omega =10\text{rad/s}$:

$$v=0.5\cdot 10=5\text{m/s}$$

$$a_r=\frac{5^2}{0.5}=50{\text{m/s}}^2$$

Answer: $v=5\text{m/s}$, $a_r=50{\text{m/s}}^2$

Projectile motion

Projectile motion describes the two-dimensional motion of a particle under gravity alone (air resistance neglected). The key idea is that the motion separates into:

• horizontal motion with constant velocity
• vertical motion with constant acceleration $-g$

From this, you can find quantities like time of flight, range, and maximum height.

• Horizontal:

  $$x=v_0\cos \theta \cdot t$$

• Vertical:

  $$y=v_0\sin \theta \cdot t-\frac{1}{2}g{t}^2$$

• Time of flight:

  $$T=\frac{2v_0\sin \theta }{g}$$

• Range:

  $$R=\frac{v_0^2\sin 2\theta }{g}$$

Example: range of a projectile A projectile is launched with $v_0=15\text{m/s}$ at $\theta =45^{\circ }$:

$$R=\frac{15^2\sin (90^{\circ })}{9.81}\approx 22.9\text{m}$$

Answer: $R\approx 22.9\text{m}$

Particle kinetics (Newton’s second law)

Particle kinetics connects motion to the forces that cause it. Newton’s Second Law states that the net force on a particle equals mass times acceleration. You can use this relationship in two common ways:

• find acceleration when forces are known
• find required forces when the motion is specified

$$\sum \vec{F}=m\vec{a}$$

Example: acceleration from an applied force A block of mass $8\text{kg}$ is pulled by a force $40\text{N}$. Find the acceleration.

$$a=\frac{F}{m}=\frac{40}{8}=5{\text{m/s}}^2$$

Answer: $a=5{\text{m/s}}^2$

Principle of work and energy

The principle of work and energy relates forces and motion through energy rather than acceleration. It states that the work done by all forces as a particle moves from position 1 to position 2 equals the change in kinetic energy.

This approach is especially useful when forces depend on position, because it avoids solving directly for acceleration as a function of time.

$$T_1+U_{1\to 2}=T_2$$

• Kinetic energy:

  $$T=\frac{1}{2}m{v}^2$$

• Work:

  $$U=\int \vec{F}\cdot d\vec{r}$$

Example: final velocity from work and energy A $4\text{kg}$ block initially at rest is acted upon by a constant force $F=12\text{N}$ through $3\text{m}$. Find the final velocity.

$$U=12\cdot 3=36\text{J}$$

The block starts from rest, so $T_1=0$ and $T_2=36\text{J}$.

$$T=\frac{1}{2}(4)v^2=36\Rightarrow v=\sqrt{18}\approx 4.24\text{m/s}$$

Answer: $v\approx 4.24\text{m/s}$

Kinetic and potential energy

Kinetic energy is energy of motion. Potential energy is stored energy associated with position or configuration in a force field.

In many mechanical problems, you’ll work with:

• gravitational potential energy
• elastic (spring) potential energy

Mechanical energy is conserved when only conservative forces act.

• Particle KE:

  $$T=\frac{1}{2}m{v}^2$$

• Rigid body KE (rotation):

  $$T=\frac{1}{2}I{\omega }^2$$

• Gravitational PE:

  $$V=mgh$$

Example: gravitational potential energy gained A $2\text{kg}$ mass is lifted $h=5\text{m}$. Potential energy gained:

$$V=2\cdot 9.81\cdot 5=98.1\text{J}$$

Answer: $V=98.1\text{J}$

Work, power, and efficiency

Work measures energy transfer due to a force acting through a displacement. Power measures how quickly work is done. Efficiency compares useful output to total input.

• Work:

  $$W=\int \vec{F}\cdot d\vec{r}$$

• Power:

  $$P=\frac{dW}{dt}=\vec{F}\cdot \vec{v}$$

• Efficiency:

  $$\eta =\frac{\text{Output}}{\text{Input}}\times 100\%$$

Example: motor efficiency A motor delivers $500\text{J}$ of work input with $400\text{J}$ output.

$$\eta =\frac{400}{500}\times 100\%=80\%$$

Answer: $\eta =80\%$

Impulse and momentum

Impulse and momentum methods connect force and motion over a time interval. They’re especially useful when forces act for a short time (such as during a hit or collision) or when the force varies with time.

• Linear impulse:

  $$\vec{J}=\int \vec{F}dt$$

• Linear momentum:

  $$\vec{p}=m\vec{v}$$

• Impulse-momentum principle:

  $$\vec{J}=\Delta \vec{p}$$

Example: velocity change from an impulsive force A $2\text{kg}$ ball is struck with force $F=50\text{N}$ for $0.1\text{s}$:

$$J=50\cdot 0.1=5\text{Ns}$$

Velocity gained:

$$\Delta v=\frac{J}{m}=\frac{5}{2}=2.5\text{m/s}$$

Answer: $\Delta v=2.5\text{m/s}$

Angular momentum

Angular momentum is the rotational counterpart of linear momentum. It can be defined about a point, and its rate of change depends on the net external moment about that point.

When external moments are negligible, angular momentum is conserved.

• Angular momentum:

  $$H=I\omega$$

• Angular impulse:

  $$\int \vec{M}dt=\Delta H$$

Example: angular momentum of a flywheel A flywheel with $I=4{kg\cdot m}^2$ spins at $\omega =20\text{rad/s}$:

$$H=4\cdot 20=80{kg\cdot m}^2/\text{s}$$

Answer: $H=80{kg\cdot m}^2/\text{s}$

Momentum conservation is a special case of this impulse-momentum relationship, and it applies to angular momentum the same way:

• Linear: if $\sum {\vec{F}}_{\text{ext}}=0$, then ${\vec{p}}_{\text{initial}}={\vec{p}}_{\text{final}}$
• Angular: if $\sum M=0$, then $H_{\text{initial}}=H_{\text{final}}$

Example: conservation of momentum in an elastic collision A $3\text{kg}$ cart moving at $2\text{m/s}$ collides elastically with a $2\text{kg}$ cart at rest. Find the total momentum and the final velocity of each cart.

$$p=3\cdot 2=6kg\cdot m/s$$

Final total momentum must also equal $6kg\cdot m/s$: $3v_1^{\prime }+2v_2^{\prime }=6$. Because the collision is elastic, $e=1$, so $v_2^{\prime }-v_1^{\prime }=2\text{m/s}$. Solving the two equations together gives $v_1^{\prime }=0.4\text{m/s}$ and $v_2^{\prime }=2.4\text{m/s}$.

Answer: $v_1^{\prime }=0.4\text{m/s}$, $v_2^{\prime }=2.4\text{m/s}$

Impact

Impact is a collision that occurs over a short time interval, producing large impulsive forces. During the collision, external forces are often treated as negligible compared with the impulsive contact forces, so momentum methods are commonly used.

Collisions are classified by how much kinetic energy is lost:

• elastic
• partially elastic
• perfectly inelastic

The coefficient of restitution measures how “elastic” the collision is.

• Coefficient of restitution:

  $$e=\frac{v_{B2}-v_{A2}}{v_{A1}-v_{B1}},0\le e\le 1$$

Example: coefficient of restitution Two balls collide: $v_{A1}=4\text{m/s}$, $v_{B1}=0$, $v_{A2}=1\text{m/s}$, $v_{B2}=?$, $e=0.8$. In general, finding both post-impact velocities requires pairing this equation with conservation of momentum, but here $v_{A2}$ is already known, so the restitution equation alone solves for $v_{B2}$.

$$0.8=\frac{v_{B2}-1}{4-0}\Rightarrow v_{B2}=4.2\text{m/s}$$

Answer: $v_{B2}=4.2\text{m/s}$

Dynamic friction

Dynamic (kinetic) friction is the resistive force between two surfaces sliding relative to each other. A common model assumes the friction magnitude is proportional to the normal force.

Watch out: these equations assume a consistent unit system. In SI, use $g=9.81{\text{m/s}}^2$ with mass in kilograms. In USCS, use $g=32.2{\text{ft/s}}^2$ with mass in slugs (recall $1\text{slug}=32.2\text{lbm}$) - mixing lbm and lbf without converting through $g_c$ is a common FE trap. Also carry intermediate results at full calculator precision, since FE distractors are often spaced closely enough to catch an early rounding.

• Kinetic friction force:

  $$f_k={\mu }_{k}N$$

Example: kinetic friction force A $10\text{kg}$ block slides on a surface with ${\mu }_k=0.3$.

$$f_k=0.3\cdot (10\cdot 9.81)\approx 29.4\text{N}$$

Answer: $f_k\approx 29.4\text{N}$

Plane motion of a rigid body

Plane motion of a rigid body combines translation and rotation in a single plane. Unlike particle motion, rigid body motion depends on how mass is distributed, which is captured through rotational inertia.

Common analysis tools include Newton-Euler equations and energy methods.

• Velocity of a point B on a rotating body:

  $${\vec{v}}_B={\vec{v}}_A+\vec{\omega }\times {\vec{r}}_{B/A}$$

• Equations of motion:

  $$\sum \vec{F}=m{\vec{a}}_G$$

  $$\sum {\vec{M}}_G=I_G\alpha$$

Example: velocity of a rotating rod’s tip A rod of length $2\text{m}$ rotates about its end with $\omega =5\text{rad/s}$. Find the velocity of the tip.

$$v=r\omega =2\cdot 5=10\text{m/s}$$

Answer: $v=10\text{m/s}$

Free and forced vibration

Vibration is oscillatory motion. The two basic cases are:

• Free vibration: motion caused by an initial disturbance, with no continuing external excitation
• Forced vibration: motion driven by an ongoing external load

System response depends on properties such as mass, stiffness, and damping. Natural frequency and damping ratio are commonly used to describe that response.

Free vibration (undamped)

$$m\ddot{x}+kx=0$$

General solution:

$$x(t)=A\cos ({\omega }_{n}t)+B\sin ({\omega }_{n}t)$$

where

$${\omega }_n=\sqrt{\frac{k}{m}}$$

Example: natural frequency For $m=1\text{kg}$, $k=25\text{N/m}$:

$${\omega }_n=\sqrt{\frac{25}{1}}=5\text{rad/s}$$

Answer: ${\omega }_n=5\text{rad/s}$

Forced vibration

$$m\ddot{x}+kx=F_0\cos (\omega t)$$

The steady-state response oscillates at the forcing frequency with amplitude $X=\frac{F_0/k}{1-(\omega /{\omega }_n{)}^2}$. As $\omega \to {\omega }_n$, $X$ grows without bound, a condition called resonance - which is why designers keep operating frequencies away from a system’s natural frequency.

Example: steady-state amplitude For $m=2\text{kg}$, $k=50\text{N/m}$, $F_0=10\text{N}$, $\omega =3\text{rad/s}$ (so ${\omega }_n=\sqrt{50/2}=5\text{rad/s}$):

$$X=\frac{10/50}{1-(3/5{)}^2}=\frac{0.2}{0.64}\approx 0.313\text{m}$$

Answer: $X\approx 0.313\text{m}$

Torsional vibration

Torsional vibration is angular oscillation about a system’s longitudinal axis. It commonly appears in rotating shafts, drive trains, and power transmission systems.

The equations mirror linear vibration, but use angular displacement, torsional stiffness, and mass moment of inertia.

• Equation of motion:

  $$I\ddot{\theta }+k_{\theta }\theta =0$$

• Natural frequency:

  $${\omega }_n=\sqrt{\frac{k_{\theta }}{I}}$$

Example: natural frequency of torsional vibration For $I=0.5{kg\cdot m}^2$, $k_{\theta }=200N\cdot m/rad$:

$${\omega }_n=\sqrt{\frac{200}{0.5}}=20\text{rad/s}$$

Answer: ${\omega }_n=20\text{rad/s}$