9 - Rotational Motion and Dynamics

A structured guide to angular motion, torque, rotational inertia, rotational dynamics, rolling, and equilibrium about a fixed axis.

The rotational framework

Rotational motion describes an object turning about an axis. For a rigid object, the shape remains effectively unchanged, so each point completes the same angular motion even though points at different distances from the axis can have different linear speeds.

The main rotational quantities correspond closely to translational quantities:

Angular position $\theta$ corresponds to position $x$ .
$\omega$ corresponds to velocity $v$ .
$\alpha$ corresponds to acceleration $a$ .
$I$ corresponds to mass $m$ .
$\tau$ corresponds to force $F$ .

The central rotational-dynamics relationship is

\sum\tau=I\alpha.

Takeaway: Rotational problems become easier when each angular quantity is matched with its translational analog, while remembering that and replace force and mass.

Angular position and displacement

Angular position $\theta$ specifies an object’s orientation relative to a reference direction. Angular displacement is

\Delta\theta=\theta_f-\theta_i.

Use radians for rotational equations. One complete revolution is

1\text{ revolution}=2\pi\text{ rad}=360^\circ.

If a point lies a perpendicular distance $r$ from the axis and the object rotates through an angle $\theta$ , the point travels an arc length

s=r\theta.

For example, a point on a wheel of radius $0.40\,\text{m}$ moving through $90^\circ=\frac{\pi}{2}\,\text{rad}$ travels

s=(0.40)\left(\frac{\pi}{2}\right)\approx1.26\,\text{m}.

Choose a sign convention, commonly counterclockwise positive and clockwise negative, and use it consistently.

Takeaway: Convert degrees to radians before using $s=r\theta$ or rotational kinematic equations.

and acceleration

describes the rate of change of angular position:

\omega_{\text{avg}}=\frac{\Delta\theta}{\Delta t}.

The direction of the angular-velocity vector lies along the rotation axis and can be found with the right-hand rule. Curl the fingers of your right hand in the direction of rotation; your thumb points in the direction of $\vec{\omega}$ .

describes the rate of change of :

\alpha_{\text{avg}}=\frac{\Delta\omega}{\Delta t}.

For a point at radius $r$ , the tangential quantities are

v_t=r\omega

and

a_t=r\alpha.

All points on a rigid object have the same $\omega$ and $\alpha$ , but a point farther from the axis has greater tangential speed and tangential acceleration. A rotating point also has centripetal acceleration toward the axis:

a_c=\frac{v_t^2}{r}=r\omega^2.

Tangential acceleration changes speed, whereas centripetal acceleration changes the direction of velocity. Also, positive does not always mean speeding up; the signs of $\omega$ and $\alpha$ must be compared.

Takeaway: Separate angular rates from tangential rates, and separate the roles of tangential and centripetal acceleration.

Rotational kinematics

When is constant, rotational kinematics follows the same mathematical pattern as constant-acceleration linear motion:

\omega_f=\omega_i+\alpha t

\theta_f=\theta_i+\omega_i t+\frac{1}{2}\alpha t^2

\omega_f^2=\omega_i^2+2\alpha\Delta\theta

\Delta\theta=\frac{\omega_i+\omega_f}{2}t.

These equations require constant $\alpha$ , and angular displacements must be measured in radians.

For a wheel starting from rest with $\alpha=3.0\,\text{rad/s}^2$ for $4.0\,\text{s}$ ,

\omega_f=0+(3.0)(4.0)=12\,\text{rad/s}

and

\Delta\theta=0+\frac{1}{2}(3.0)(4.0)^2=24\,\text{rad}.

The number of revolutions is $\frac{24}{2\pi}\approx3.8$ .

Takeaway: Select the kinematic equation that contains the known quantities and the desired unknown, and verify that the is constant.

and lever arms

depends on the force, its distance from the axis, and its direction:

\tau=rF\sin\phi.

Here, $\phi$ is the angle between the position vector and the force. The equivalent lever-arm form is

\lvert\tau\rvert=r_\perp F,

where the lever arm is the perpendicular distance from the axis to the force’s line of action.

A perpendicular force has maximum because $\sin 90^\circ=1$ . A radial force has zero because $\sin 0^\circ=0$ . For example, a perpendicular force of $30\,\text{N}$ applied $0.80\,\text{m}$ from a door’s hinges produces

\tau=(0.80)(30)=24\,\text{N}\cdot\text{m}.

Assign signs to torques according to the selected convention, then add signed values to obtain net . Do not add opposing magnitudes as though they acted in the same direction.

has units of $\text{N}\cdot\text{m}$ , which are dimensionally the same as joules, but and energy are different physical quantities.

Takeaway: Use the perpendicular distance and the force direction, not merely the distance to the point where the force is applied.

Rotational dynamics

measures resistance to and depends on the mass distribution relative to the selected axis. For point masses,

I=\sum_i m_i r_i^2.

Because distance is squared, moving mass farther from the axis can substantially increase $I$ , even when total mass is unchanged. Common idealized expressions include:

Point mass at radius $R$ : $I=mR^2$ .
Thin hoop about its center: $I=MR^2$ .
Solid disk or cylinder about its center: $I=\frac{1}{2}MR^2$ .
Slender rod about its center: $I=\frac{1}{12}ML^2$ .
Slender rod about one end: $I=\frac{1}{3}ML^2$ .

For a disk with $I=2.0\,\text{kg}\cdot\text{m}^2$ and net $6.0\,\text{N}\cdot\text{m}$ , gives

\alpha=\frac{\sum\tau}{I}=\frac{6.0}{2.0}=3.0\,\text{rad/s}^2.

A reliable solution process is:

Choose and state the axis.
Identify all forces that produce about that axis.
Assign signs to the torques.
Calculate the net .
Determine about the same axis.
Apply $\sum\tau=I\alpha$ .
Use rotational kinematics if , position, or time is requested.

Takeaway: The same axis must be used for every and for the .

Rolling and equilibrium

Many systems translate and rotate at the same time. For an object , the center-of-mass motion and rotation are connected by

v_{\text{cm}}=R\omega

and

a_{\text{cm}}=R\alpha.

These relations describe the instantaneous motion at the contact point and do not automatically apply when the object slides.

For combined motion, analyze two linked parts:

Forces determine the translational acceleration of the center of mass.
Torques determine about the center of mass or another chosen axis.

An object is in when

\alpha=0,

which requires

\sum\tau=0.

If it is also in translational equilibrium, then

\sum F=0.

For two perpendicular forces balancing a lever,

r_1F_1=r_2F_2.

Thus, a smaller force can balance a larger force when it acts farther from the axis.

Takeaway: Combined motion may require both force and equations; equilibrium requires the appropriate net quantities to vanish.