Unit 5: Rotation - Comprehensive Guide

5.1 Torque and Rotational Statics

Defining Torque

Torque ($\tau$) is the rotational equivalent of force. While force causes linear acceleration, torque causes angular acceleration. It defines how effective a force is at twisting an object around an axis.

Technically, torque is a vector quantity defined by the cross product of the position vector (from the axis or pivot to the point of force application) and the force vector.

$\vec{\tau} = \vec{r} \times \vec{F}$

The magnitude of torque is given by:

$\tau = rF\sin(\theta)$

Where:

• $r$ is the distance from the pivot to the point of force application.
• $F$ is the magnitude of the force.
• $\theta$ is the angle between the position vector $\vec{r}$ and the force vector $\vec{F}$.

Alternatively, torque can be thought of as the force multiplied by the moment arm ($r_{\perp}$ or lever arm). The moment arm is the perpendicular distance from the axis of rotation to the "line of action" of the force.

$\tau = r_{\perp} F$

Direction: The Right-Hand Rule

Since torque is a vector, it has direction along the axis of rotation.

• Counter-clockwise torques are generally considered positive (+).
• Clockwise torques are generally considered negative (-).

To find the vector direction: Point your fingers in the direction of $\vec{r}$, curl them toward $\vec{F}$, and your thumb points in the direction of $\vec{\tau}$ (out of the page or into the page).

Rotational Equilibrium

For a rigid body to be in static equilibrium, two conditions must be met:

1. Translational Equilibrium: The net external force is zero ($\Sigma \vec{F} = 0$).
2. Rotational Equilibrium: The net external torque is zero ($\Sigma \vec{\tau} = 0$) about any pivot point.

Tip: In static equilibrium problems (like a ladder leaning against a wall), you can choose your pivot point anywhere. Always choose a pivot where unknown forces act (like a hinge) to eliminate their torque from the equation ($r=0$, so $\tau=0$).

5.2 Rotational Kinematics

Variables and Analogies

Rotational motion creates a direct parallel to translational (linear) motion. All points on a rigid rotating body move through the same angle in the same amount of time.

Note: $\theta$ must be in radians for the relationship equations to hold true.

The "Big Five" for Constant Angular Acceleration

If angular acceleration ($\alpha$) is constant, the kinematic equations are identical in form to linear kinematics.

1. $\omega<em>f = \omega</em>i + \alpha t$
2. $\Delta \theta = \frac{1}{2}(\omega<em>i + \omega</em>f)t$
3. $\Delta \theta = \omega_i t + \frac{1}{2}\alpha t^2$
4. $\omega<em>f^2 = \omega</em>i^2 + 2\alpha \Delta \theta$
5. $\Delta \theta = \omega_f t - \frac{1}{2}\alpha t^2$

Non-Uniform Acceleration

In AP Physics C, $\alpha$ is not always constant. You must recall the calculus definitions:

$\omega = \frac{d\theta}{dt} \quad \text{and} \quad \alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$

5.3 Rotational Inertia (Moment of Inertia)

Rotational Inertia ($I$), often called the Moment of Inertia, is the measure of an object's resistance to changes in its rotational motion. It is the rotational analog of mass.

Discrete Particles

For a system of distinct point masses:

$I = \sum m<em>i r</em>i^2$

Where $ri$ is the perpendicular distance of mass $mi$ from the axis of rotation.

Continuous Objects (Calculus Approach)

For rigid, continuous bodies, the summation becomes an integral over the volume of the object:

$I = \int r^2 dm$

To solve this, you typically express $dm$ in terms of spatial dimensions using density:

• Linear: $\lambda = \frac{M}{L} \Rightarrow dm = \lambda dr$
• Area: $\sigma = \frac{M}{A} \Rightarrow dm = \sigma dA$
• Volume: $\rho = \frac{M}{V} \Rightarrow dm = \rho dV$

Example: Thin Rod Rotated About End
Consider a rod of mass $M$ and length $L$. Density $\lambda = M/L$.
$I = \int<em>{0}^{L} x^2 (\lambda dx) = \frac{M}{L} \left[ \frac{x^3}{3} \right]</em>0^L = \frac{M}{L}\frac{L^3}{3} = \frac{1}{3}ML^2$

Parallel Axis Theorem

If you know the moment of inertia about the center of mass ($I_{cm}$), you can calculate the inertia about any parallel axis shift by distance $d$:

$I = I_{cm} + Md^2$

Common $I_{cm}$ values to memorize:

• Hoop/Cylindrical Shell: $MR^2$
• Solid Cylinder/Disk: $\frac{1}{2}MR^2$
• Solid Sphere: $\frac{2}{5}MR^2$
• Rod (center): $\frac{1}{12}ML^2$

5.4 Rotational Dynamics

Newton's Second Law for Rotation

Just as $F_{net} = ma$, the angular acceleration of a rigid body is proportional to the net torque and inversely proportional to the rotational inertia:

$\Sigma \tau = I\alpha$

Strategy for Dynamics Problems

1. Draw a Free Body Diagram (FBD) showing forces at their points of application.
2. Identify the axis of rotation.
3. Write $\Sigma F = ma$ for the center of mass (translational motion).
4. Write $\Sigma \tau = I\alpha$ about the center of mass (rotational motion).
5. Link the motions: If the object rolls without slipping or a rope unwinds without slipping, equate $a = R\alpha$.

5.5 Energy and Rolling Motion

Rotational Kinetic Energy

A rotating object possesses kinetic energy due to its motion:

$K_{rot} = \frac{1}{2}I\omega^2$

Rolling Motion (Rolling Without Slipping)

Rolling is a combination of translation of the center of mass and rotation about the center of mass.

Total Kinetic Energy:
$K<em>{total} = K</em>{trans} + K<em>{rot} = \frac{1}{2}Mv</em>{cm}^2 + \frac{1}{2}I_{cm}\omega^2$

The "No-Slip" Condition:
If an object rolls without slipping:

1. $v_{cm} = R\omega$
2. $a_{cm} = R\alpha$
3. The point of contact with the ground has an instantaneous velocity of zero relevant to the ground.

Crucial Concept: Friction in Rolling:
Static friction is required to cause rotation (it provides the torque). However, because the contact point does not slide (distance $d=0$ at the contact patch), static friction does no work on a rolling object. Therefore, mechanical energy is conserved for an object rolling without slipping on a ramp.

5.6 Angular Momentum

Definitions

1. For a Point Particle:
Angular momentum $\vec{L}$ is defined relative to an origin.
$\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$
Magnitude: $L = mvr\sin\theta = mvr_{\perp}$

2. For a Rigid Body:
For an object rotating about a fixed axis:
$L = I\omega$

Conservation of Angular Momentum

The rotational analog to impulse-momentum is:
$\Sigma \tau_{ext} = \frac{dL}{dt}$

Therefore, if the net external torque on a system is zero, the total angular momentum is constant.

$L<em>i = L</em>f$

Classic Examples:

• Ice Skater: Arms pull in $\rightarrow$ $I$ decreases $\rightarrow$ $\omega$ increases to conserve $L$.
• Satellite: A planet in orbit moves faster at perihelion (closest approach) because $r$ decreases, so $v$ must increase to keep $L = mvr$ constant.
• Collisions: A bullet hitting a pivoted rod. Conservation of linear momentum does not apply (pivot exerts force), but conservation of angular momentum applies about the pivot point (pivot exerts no torque).

Sample Problem: The Merry-Go-Round

A child of mass $m$ runs with velocity $v$ tangent to the rim of a merry-go-round (disk of mass $M$, radius $R$) and jumps on.

Solution:
Consider the axis of rotation at the center. No external torques act about this axis.

1. Initial $L$: Only the child has angular momentum. Treat child as a point particle.
   $L<em>i = R \times p</em>{child} = R(mv)$
2. Final $L$: The child and disk rotate together.
   $I<em>{total} = I</em>{disk} + I<em>{child} = \frac{1}{2}MR^2 + mR^2$$L</em>f = I<em>{total}\omega</em>f$
3. Conservation:
   $mRv = (\frac{1}{2}MR^2 + mR^2)\omega<em>f$$\omega</em>f = \frac{mRv}{0.5MR^2 + mR^2}$

Common Mistakes & Pitfalls

1. Mixing Degrees and Radians: All rotational kinematic formulas ($s=r\theta$, etc.) require radians. Never use degrees.
2. Confusing $\alpha$ and $a$: Remember $a = r\alpha$ refers to the tangential acceleration. Rotating objects also have centripetal acceleration ($a_c = v^2/r = \omega^2 r$) directed inward.
3. Forgeting the Axis Theorem: Students often memorize $I = \frac{1}{12}ML^2$ for a rod and try to use it when the rod rotates around its end. You must use the Parallel Axis Theorem ($I_{end} = \frac{1}{3}ML^2$).
4. Conservation of Momentum Confusion: In collision problems involving rotation (like a clay ball hitting a hanging stick), Linear Momentum is usually NOT conserved (due to the hinge force), considering the system of ball+stick. Only Angular Momentum is conserved about the hinge.
5. Torque vs. Force: A net force of zero does not mean net torque is zero (and vice versa). A "force couple" (two equal opposite forces separated by distance) creates torque but zero net force.