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.

τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}

The magnitude of torque is given by:

τ=rFsin⁡(θ)\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.

τ=r⊥F\tau = r_{\perp} F

Diagram showing a wrench turning a nut. Vectors for force F, radius r, and the angle theta are labeled. A dashed line extends the force vector to show the perpendicular moment arm r_perp.

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.

Linear VariableSymbolRotational VariableSymbolRelationship to Linear
Displacement$x$Angular Displacement$\theta$$x = r\theta$
Velocity$v$Angular Velocity$\omega$$v = r\omega$
Acceleration$a$Angular Acceleration$\alpha$$a_t = r\alpha$

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. ω<em>f=ω</em>i+αt\omega<em>f = \omega</em>i + \alpha t
  2. Δθ=12(ω<em>i+ω</em>f)t\Delta \theta = \frac{1}{2}(\omega<em>i + \omega</em>f)t
  3. Δθ=ωit+12αt2\Delta \theta = \omega_i t + \frac{1}{2}\alpha t^2
  4. ω<em>f2=ω</em>i2+2αΔθ\omega<em>f^2 = \omega</em>i^2 + 2\alpha \Delta \theta
  5. Δθ=ωft−12αt2\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:

ω=dθdtandα=dωdt=d2θdt2\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=∑m<em>ir</em>i2I = \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=∫r2dmI = \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=∫<em>0Lx2(λdx)=ML[x33]</em>0L=MLL33=13ML2I = \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

Calculus derivation diagram of a rod. Shows a small slice 'dx' at distance 'x' from the axis. The integral sum is visualized.

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=Icm+Md2I = 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:

Στ=Iα\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:

Krot=12Iω2K_{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=12Mv</em>cm2+12Icmω2K<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.

Diagram of rolling motion breakdown. Part A: Pure translation (velocity v everywhere). Part B: Pure rotation (top has v, bottom has -v). Part C: Combined Rolling (Top has 2v, Center has v, Bottom contact point has 0 velocity).

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.
L⃗=r⃗×p⃗=m(r⃗×v⃗)\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=IωL = I\omega

Conservation of Angular Momentum

The rotational analog to impulse-momentum is:
Στext=dLdt\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>fL<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×p</em>child=R(mv)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=12MR2+mR2I<em>{total} = I</em>{disk} + I<em>{child} = \frac{1}{2}MR^2 + mR^2L</em>f=I<em>totalω</em>fL</em>f = I<em>{total}\omega</em>f
  3. Conservation:
    mRv=(12MR2+mR2)ω<em>fmRv = (\frac{1}{2}MR^2 + mR^2)\omega<em>fω</em>f=mRv0.5MR2+mR2\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.