Rotational Mechanics

Apparent Paradox:

• A body can have zero resultant external force (ΣF = 0)
• Yet still not be in rotational equilibrium

Resolution:

• Translational equilibrium requires: ΣF = 0
• Rotational equilibrium requires: ΣΓ = 0 (zero net torque)
• These are independent conditions

Example: A pulley with a bucket going down a well

• Net force on pulley is zero (center of mass stationary)
• But torque is non-zero (causing angular acceleration)

For complete equilibrium, both conditions must be satisfied independently:

1. ΣF = 0 (no linear acceleration)
2. ΣΓ = 0 (no angular acceleration)

Three Types of Equilibrium:

1. Stable Equilibrium:

   • When slightly displaced and released, the body returns to its original position
   • Center of mass rises when displaced
   • Potential energy is at a local minimum
   • Example: A ball in a bowl

2. Unstable Equilibrium:

   • When slightly displaced and released, the body moves further away
   • Center of mass falls when displaced
   • Potential energy is at a local maximum
   • Example: A pencil balanced on its tip

3. Neutral Equilibrium:

   • When slightly displaced and released, the body remains in the new position
   • Center of mass height remains unchanged when displaced
   • Potential energy remains constant
   • Example: A ball on a flat surface

Determining type: Examine what happens to the center of mass height during a small displacement. The direction of potential energy change (increase, decrease, or constant) indicates the type of equilibrium.

• Constraints on degrees of freedom:

  • A rigid body normally has 6 degrees of freedom (3 translational, 3 rotational)
  • Fixing an axis reduces this to just 1 degree of freedom (rotation about that axis)
  • All possible displacements are restricted to rotations about this single axis

• Governing principle:

  • The rigid body constraint ensures all particles maintain fixed distances from each other
  • The rotation constraint ensures all particles move in circles with centers on the axis
  • Motion is described by a single variable: the angular position $\theta$
  • Dynamics is governed by $\Gamma = I\alpha$, where $\Gamma$ is torque and $I$ is moment of inertia

Example: When you hold a tennis racket at both ends and rotate it, despite its complex shape, its motion is completely described by a single angle of rotation about the axis between your hands.

A force produces zero torque about an axis in two specific cases:

1. When the force is parallel to the axis of rotation

   • In this case, the cross product $\vec{r} \times \vec{F}$ is perpendicular to the axis, so its component along the axis is zero

2. When the force intersects the axis of rotation

   • Taking the point of intersection as the origin, the position vector $\vec{r}$ and force $\vec{F}$ lie along the same line
   • This makes their cross product zero: $\vec{r} \times \vec{F} = 0$

Any force that neither intersects nor is parallel to the axis will produce a non-zero torque about that axis.

To determine the torque about axis AB:

1. Find the plane that passes through point P and is perpendicular to the axis AB
2. Locate point O where this plane intersects the axis
3. Draw the position vector $\vec{r} = \overrightarrow{OP}$ from O to P
4. Calculate the torque using: $\tau = |\vec{r} \times \vec{F}| = F \cdot r \sin\theta$ Where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$

Alternatively:

• Find the common perpendicular (shortest distance) from the axis to the line of action of the force
• Multiply the force magnitude by this distance: $\tau = F \cdot d$

The direction of the torque is along the axis, determined by the right-hand rule.

Physical significance of force components in deriving Γ = Iα:

1. Radial component (mω²r):

• Acts along the radius toward the axis
• Creates no torque about the rotation axis because its line of action passes through or is parallel to the axis
• Maintains circular path but cannot change rotational speed
• Forms an internal constraint force within the rigid body system

2. Tangential component (mrα):

• Acts perpendicular to the radius
• Produces torque mr²α about the rotation axis
• Directly responsible for angular acceleration
• Summing these torques for all particles gives Γ = (Σmr²)α = Iα

This separation reveals the fundamental principle that only forces with components perpendicular to the radius can change a body's rotation rate, while radial forces merely maintain the circular path. This parallels how only forces parallel to velocity can change speed in linear motion.

The total angular momentum of a rigid body rotating about axis AB is the sum of the angular momenta of all its constituent particles:

L = ∑mᵢvᵢrᵢ = ∑mᵢ(rᵢω)rᵢ = ω∑mᵢrᵢ² = Iω

Where:

• mᵢ = mass of each particle
• rᵢ = perpendicular distance of each particle from the axis
• vᵢ = tangential velocity of each particle (vᵢ = rᵢω)
• I = moment of inertia about the axis AB (I = ∑mᵢrᵢ²)
• ω = angular velocity (same for all particles in a rigid body)

This summation approach works because each particle in the rigid body moves in a circular path perpendicular to the axis of rotation, with its own contribution to the total angular momentum.

Derivation of tan(θ) = v²/rg for a leaning cyclist:

1. Choose a rotating reference frame with origin at the turn's center

2. In this frame, the cyclist is stationary but experiences a centrifugal force Mv²/r

3. For translational equilibrium:

   • Vertical: N·cos(θ) = Mg
   • Horizontal: N·sin(θ) = Mv²/r

4. Taking the ratio:

   • N·sin(θ)/N·cos(θ) = (Mv²/r)/Mg
   • tan(θ) = v²/rg

Assumptions:

• The road is not banked
• No slipping occurs (adequate friction)
• Cyclist and bicycle form a rigid body
• The center of mass is directly above the contact points
• Air resistance is negligible
• Gyroscopic effects from wheels are negligible

The formula shows that faster speeds or smaller turn radii require greater lean angles. This explains why motorcyclists lean more dramatically when taking sharp turns at high speeds.

Using conservation of angular momentum:

Initial angular momentum = Final angular momentum $I_1\omega + I_2(0) = (I_1 + I_2)\omega'$

Solving for $\omega'$: $\omega' = \frac{I_1\omega}{I_1 + I_2}$

Example: For two identical wheels ($I_1 = I_2 = I$) where one rotates at $\omega$ and the other is at rest, when coupled: $\omega' = \frac{I\omega}{2I} = \frac{\omega}{2}$

The final angular velocity always decreases due to conservation of angular momentum distributed over greater rotational inertia.

For a particle P(xᵢ, yᵢ, zᵢ):

1. Perpendicular distance to X-axis:

   • r = √(yᵢ² + zᵢ²)
   • Uses Pythagorean theorem in the Y-Z plane
   • xᵢ coordinate is irrelevant as it represents position along the axis

2. Perpendicular distance to Y-axis:

   • r = √(xᵢ² + zᵢ²)
   • Uses Pythagorean theorem in the X-Z plane
   • yᵢ coordinate doesn't contribute to distance from Y-axis

3. Perpendicular distance to Z-axis:

   • r = √(xᵢ² + yᵢ²)
   • Uses Pythagorean theorem in the X-Y plane
   • zᵢ coordinate doesn't contribute to distance from Z-axis

This relationship is fundamental for calculating moment of inertia I = ∑mᵢrᵢ², which determines an object's resistance to rotational acceleration.