Summary

General Principles

The instantaneous velocity

\vec{v} = d \vec{r} / d t

is a vector tangent to the trajectory.

The instantaneous acceleration is

\vec{a} = d \vec{v} / d t

${\vec{a}}_{∥}$ the component of $\vec{a}$ parallel to $\vec{v}$ , is responsible for change of speed. $\vec{a}_{⊥}$ , the component of $\vec{a}$ perpendicular to $\vec{v}$ , is responsible for change of direction.

The figure illustrates the parallel and perpendicular components of instantaneous acceleration of a particle.

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Relative Motion

If object C moves relative to reference frame A with velocity $\vec{v}_{CA}$ , then it moves relative to a different reference frame B with velocity

\vec{v}_{CB} = \vec{v}_{CA} + \vec{v}_{AB}

where $\vec{v}_{AB}$ is the velocity of A relative to B. This is the Galilean transformation of velocity.

The figure illustrates two x y coordinate planes that depict reference frames labeled “A” and B.

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Important Concepts

Uniform Circular Motion

Angular velocity $ω = d θ / d t$ .

$v_{t}$ and $ω$ are constant:

v_{t} = ω r

The centripetal acceleration points toward the center of the circle with magnitude

a = | \vec{a} | = \frac{v_{t}^{2}}{r} = ω^{2} r

It changes the particle’s direction but not its speed.

The figure illustrates the velocity vector and acceleration vector of an object in uniform circular motion.

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Nonuniform Circular Motion

Angular acceleration $α = d ω / d t$ .

The radial acceleration

a_{r} = \frac{v_{t}^{2}}{r} = ω^{2} r

changes the particle’s direction. The tangential component

a_{t} = α r

changes the particle’s speed.

The figure illustrates the components of the acceleration vector of an object in nonuniform circular motion.

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Applications

Kinematics in two dimensions

If $\vec{a}$ is constant, then the x- and y-components of motion are independent of each other.

\begin{array}{l} x_{f} = x_{i} + v_{i x} Δ t + \frac{1}{2} a_{x} {(Δ t)}^{2} \\ y_{f} = y_{i} + v_{i y} Δ t + \frac{1}{2} a_{y} {(Δ t)}^{2} \\ v_{f x} = v_{i x} + a_{x} Δ t \\ v_{f y} = v_{i y} + a_{y} Δ t \end{array}

Projectile motion is motion under the influence of only gravity.

Model Model as a particle launched with speed $v_{0}$ at angle $θ$ .

Visualize Use coordinates with the x-axis horizontal and the y-axis vertical.

Solve The horizontal motion is uniform with $v_{x} = v_{0} cos θ$ . The vertical motion is free fall with $a_{y} = - g$ . The x and y kinematic equations have the same value for $Δ t$ .

The figure illustrates the trajectory of a particle in projectile motion.

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Circular motion kinematics

Period $T = \frac{2 π r}{v} = \frac{2 π}{ω}$

Angular position $θ = \frac{s}{r}$

Constant angular acceleration

\begin{array}{l} ω_{f} = ω_{i} + α Δ t \\ θ_{f} = θ_{i} + ω_{i} Δ t + \frac{1}{2} α {(Δ t)}^{2} \\ ω_{f} ​^{2} = ω_{i} ​^{2} + 2 α Δ θ \end{array}

The figure illustrates an object in circular motion.

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Circular motion graphs and kinematics are analogous to linear motion with constant acceleration.

Angle, angular velocity, and angular acceleration are related graphically.

The angular velocity is the slope of the angular position graph.
The angular acceleration is the slope of the angular velocity graph.

The figure illustrates three graphs that depict the position-, velocity-, and acceleration-versus-time graphs for constant angular acceleration.

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