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Dynamics, Kinematics, and Vibrations

10% of exam

Kinematics and kinetics of particles and rigid bodies, work-energy and impulse-momentum, kinematics of mechanisms, and free and forced vibrations.

6 concepts

A. Kinematics of particles

Particle Kinematics: Constant Acceleration and Projectiles

Describe how a particle moves — rectilinear and curvilinear paths, the constant-acceleration kinematic equations, projectiles, normal/tangential and polar components, and relative motion.

Kinematics is the grammar of dynamics: it describes how a particle moves — position, velocity, acceleration — without ever asking what force is responsible. Almost every Dynamics question on the FE Mechanical exam begins here, because before you can write

\sum F = ma

you must know the acceleration, and before you can use work-energy you must relate speed to distance. Master the four constant-acceleration equations and the handful of component frames (Cartesian, normal-tangential, polar) and a large fraction of the topic becomes bookkeeping. The FE Reference Handbook's Dynamics chapter ('Particle Kinematics') lays out exactly these relations; this concept turns them into reflexes.

Mechanics of Materials

Material Properties and Processing

Fluid Mechanics

Thermodynamics

Heat Transfer

Measurements, Instrumentation, and Controls

Mechanical Design and Analysis

Velocity and acceleration as derivatives

Position is a vector

\mathbf{r}(t)

; velocity is its time derivative and acceleration is the derivative of velocity. In one dimension this is the chain that ties the three together, and the third form —

a\,ds = v\,dv

— is the one students forget, yet it is the bridge to work-energy because it eliminates time. Use it whenever a problem gives you a relationship between

a

and position

s

but no clock.

v = \dfrac{ds}{dt}, \qquad a = \dfrac{dv}{dt} = v\,\dfrac{dv}{ds}

The constant-acceleration equations

When acceleration is constant, the calculus collapses into three algebraic relations plus a time-free fourth. These are the workhorses: free fall (with

a = g = 9.81\ \text{m/s}^2

or

32.2\ \text{ft/s}^2

downward), braking cars, launched masses. Pick the equation that contains your three knowns and one unknown — if the problem never mentions time, reach immediately for the

v^2

form so you never have to solve for

t

first.

\begin{aligned} v &= v_0 + a t \\ s &= s_0 + v_0 t + \tfrac{1}{2} a t^2 \\ s &= s_0 + \tfrac{1}{2}(v_0 + v) t \\ v^2 &= v_0^2 + 2 a (s - s_0) \end{aligned}

Projectile motion

A projectile is just two constant-acceleration problems running in parallel: horizontal motion with

a_x = 0

(constant velocity) and vertical motion with

a_y = -g

. The two are linked only through the shared time

t

. The launch angle

\theta

splits the initial speed into

v_{0x} = v_0\cos\theta

and

v_{0y} = v_0\sin\theta

; solve the vertical equation for the time of an event, then feed that time into the horizontal equation for range. There is no horizontal force, so horizontal velocity never changes — this is the single most common conceptual win on these problems.

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

Normal and tangential components

On a curved path it is natural to resolve acceleration along the direction of motion (tangential) and toward the center of curvature (normal). The tangential component

a_t = dv/dt

changes the speed; the normal component

a_n = v^2/\rho

changes the direction and always points inward. Even at constant speed (

a_t = 0

) a particle on a curve accelerates, because

a_n

is nonzero — this is why a car rounding a bend needs friction. The total magnitude is the Pythagorean sum of the two.

a_t = \dfrac{dv}{dt}, \qquad a_n = \dfrac{v^2}{\rho}, \qquad a = \sqrt{a_t^2 + a_n^2}

Radial and transverse (polar) components

When motion is described by a radius

r

and angle

\theta

— a particle on a rotating arm, a cam follower — polar components are cleanest. Note the cross terms: the radial acceleration carries a

-r\dot\theta^2

(the centripetal term), and the transverse acceleration carries the

2\dot r\dot\theta

Coriolis term that appears whenever the particle moves radially while the arm rotates. Dropping that Coriolis term is a classic error on rotating-arm problems.

a_r = \ddot r - r\dot\theta^2, \qquad a_\theta = r\ddot\theta + 2\dot r\dot\theta

Relative motion

When two particles move, the position, velocity, and acceleration of

A

relative to a translating frame on

B

add vectorially:

\mathbf{v}_A = \mathbf{v}_B + \mathbf{v}_{A/B}

. Treat each as a vector — break into components, add, then recombine. Relative-velocity problems (a boat in a current, two cars at an intersection) are pure vector addition; the trap is mixing up which body is the reference, so always read the subscript order as 'of A with respect to B.'

\mathbf{r}_A = \mathbf{r}_B + \mathbf{r}_{A/B}, \quad \mathbf{v}_A = \mathbf{v}_B + \mathbf{v}_{A/B}, \quad \mathbf{a}_A = \mathbf{a}_B + \mathbf{a}_{A/B}

Exam strategy

First decide the frame: straight line

\to

constant-acceleration set; curve at known speed

\to

normal-tangential; rotating arm

\to

polar; two bodies

\to

relative-motion vectors. For projectiles, write the

x

and

y

equations separately and link them through

t

; the time to peak is

v_{0y}/g

and total flight (level ground) is twice that. If time is not asked and not given, use

v^2 = v_0^2 + 2a\,\Delta s

to skip it entirely. Keep

a_n = v^2/\rho

in mind even when speed is constant — the exam loves the 'constant speed therefore zero acceleration' trap, which is false on any curve.

Key equations

Kinematic derivatives

v = \dfrac{ds}{dt}, \quad a = \dfrac{dv}{dt} = v\,\dfrac{dv}{ds}

Defines velocity and acceleration. The third form eliminates time; use it for acceleration given as a function of position $s$ (m), $v$ in m/s, $a$ in m/s².

Constant-acceleration velocity

v = v_0 + a t

Velocity after time $t$ under constant $a$ ; $v_0$

Constant-acceleration position

s = s_0 + v_0 t + \tfrac{1}{2} a t^2

Average-velocity displacement

s = s_0 + \tfrac{1}{2}(v_0 + v) t

Time-free (velocity-position)

v^2 = v_0^2 + 2 a (s - s_0)

Projectile equations

x = v_0\cos\theta\,t, \quad y = v_0\sin\theta\,t - \tfrac{1}{2} g t^2

Normal/tangential acceleration

a_t = \dfrac{dv}{dt}, \quad a_n = \dfrac{v^2}{\rho}, \quad a = \sqrt{a_t^2 + a_n^2}

Polar (radial/transverse) acceleration

a_r = \ddot r - r\dot\theta^2, \quad a_\theta = r\ddot\theta + 2\dot r\dot\theta

Plane circular motion

v = r\omega, \quad a_n = r\omega^2 = \dfrac{v^2}{r}, \quad a_t = r\alpha

Relative velocity

\mathbf{v}_A = \mathbf{v}_B + \mathbf{v}_{A/B}

Worked examples

Projectile range and peak height

Problem. A ball is launched from ground level at $v_0 = 25\ \text{m/s}$ at $40^\circ$ above horizontal. Find the time of flight, horizontal range, and maximum height. Take $g = 9.81\ \text{m/s}^2$ .

Solution. Split the launch velocity:

v_{0x} = 25\cos 40^\circ = 19.15\ \text{m/s}

,

v_{0y} = 25\sin 40^\circ = 16.07\ \text{m/s}

R = \dfrac{v_0^2 \sin 2\theta}{g} = \dfrac{25^2 \sin 80^\circ}{9.81} = 62.7\ \text{m}

Accelerating car: final speed and time

Problem. A car moving at $15\ \text{m/s}$ accelerates uniformly at $2.5\ \text{m/s}^2$ over a distance of $80\ \text{m}$ . Find the final speed and the elapsed time.

Solution. No time is given, so use the time-free equation first:

v^2 = v_0^2 + 2a\,\Delta s = 15^2 + 2(2.5)(80) = 225 + 400 = 625

Total acceleration on a curve

Problem. A car travels at $20\ \text{m/s}$ on a circular track of radius $150\ \text{m}$ while speeding up at $3.0\ \text{m/s}^2$ . Find the magnitude of its total acceleration.

Solution. Normal (centripetal) component:

a_n = v^2/\rho = 20^2/150 = 2.67\ \text{m/s}^2

Common pitfalls

•Assuming constant speed means zero acceleration. On any curve $a_n = v^2/\rho \ne 0$ even when $a_t = 0$ — direction is changing.
•Forgetting horizontal velocity is constant in projectiles. There is no horizontal force, so $v_x = v_0\cos\theta$ throughout; never apply gravity to the $x$ equation.
•Mixing degrees and radians. Angular formulas ( $v = r\omega$ , $a_n = r\omega^2$ ) require $\omega$
•Dropping the Coriolis term $2\dot r\dot\theta$ in $a_\theta$ when a particle slides along a rotating arm. It is present whenever both $\dot r$
•Using the $v^2 = v_0^2 + 2a\Delta s$ form with non-constant acceleration. The constant-acceleration set is only valid when $a$ truly does not change; otherwise integrate.
•Reversing relative-motion subscripts. $\mathbf{v}_{A/B}$ is 'A relative to B'; $\mathbf{v}_A = \mathbf{v}_B + \mathbf{v}_{A/B}$
•Sign errors on vertical projectile motion. Pick a sign convention (up positive) and apply $a_y = -g$ consistently to both velocity and position.

References

NCEES FE Reference Handbook — Dynamics
Hibbeler, Engineering Mechanics: Dynamics — particle kinematics chapters

C. Newton’s second law for particles

Particle Kinetics: Newton’s Second Law, Work–Energy and Impulse–Momentum

Relate forces to motion three ways — Newton’s second law with friction, the work–energy theorem with conservative forces, and linear impulse–momentum with impact.