Rotational Mechanics

• The axis of rotation is the straight line connecting the two fixed points where the rigid body is held
• During rotation:
  • Each particle of the body moves in a circle
  • The center of each circle lies on the axis of rotation
  • The plane of each circle is perpendicular to the axis
  • Particles on the axis remain stationary
  • Particles farther from the axis trace larger circles
  • All particles complete their circular paths in the same time period

Example: When opening a door, the vertical line through the hinges forms the axis of rotation, with all points on the door moving in horizontal circles around this axis.

Analysis of motion:

• The center of mass will not accelerate because ΣF = 0 means a_CM = 0
• However, the body will experience angular acceleration α = ΣΓ/I about an axis
• The body will undergo pure rotational motion

Constraints on forces:

• Forces must be arranged so they sum to zero (ΣF = 0)
• Forces must be positioned to create non-zero net torque (ΣΓ ≠ 0)
• This requires at least two forces that are:
  1. Equal in magnitude
  2. Opposite in direction
  3. Acting along different lines (not collinear)

Example: A pulley with opposing tensions that create a couple, or a door pushed at its edge while hinged at the opposite side.

This illustrates the independence of translational and rotational dynamics.

Analysis of pulley-bucket system:

1. Angular acceleration of pulley:

   • Linear and angular acceleration are related by: a = αR
   • Therefore: α = a/R (rad/s²)

2. Tension in the rope:

   • Apply Newton's 2nd law to bucket: T - mg = -ma
   • Therefore: T = m(g-a)
   • Note: Tension is less than mg when bucket accelerates downward

3. Torque on pulley:

   • Torque = Tension × Radius = TR
   • Substituting: Γ = m(g-a)R
   • This torque equals Iα, where I is the moment of inertia of the pulley

Key insight: When the bucket accelerates downward, the pulley must correspondingly accelerate angularly to maintain the constraint that the rope doesn't slip on the pulley.

Case II in torque analysis occurs when:

• A force intersects the axis of rotation
• The force and the axis meet at a specific point

Mathematical properties:

• Torque = r⃗ × F⃗ = 0 (regardless of force magnitude)
• When using the intersection point as origin, r⃗ and F⃗ become collinear
• Therefore |r⃗ × F⃗| = |r||F|sin(0°) = 0

This differs from:

• Forces parallel to the axis (torque = 0, but vectors not collinear)
• Forces perpendicular to but not intersecting the axis (torque ≠ 0)
• General skew forces (require component analysis)

Example: Pushing a lever directly toward its pivot point produces no rotational effect, regardless of how hard you push.

A force contributes to angular acceleration around an axis if and only if:

1. It produces non-zero torque about that axis:

   • τ = r × F ≠ 0

2. Force configuration analysis:

   • Forces intersecting the axis produce zero torque
   • Forces parallel to the axis produce zero torque
   • Forces with moment arms relative to the axis produce non-zero torque

3. Mathematical test:

   • Decompose the force into components
   • Only components perpendicular to both the position vector and axis contribute
   • τ = r·F·sin(θ) where θ is angle between r and F

4. Effective torque calculation:

   • τ = F·d where d is the perpendicular distance (moment arm) from the axis to the line of action of force

Example: When turning a wrench, only the component of force perpendicular to the wrench handle contributes to rotation.

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.

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.

For a rigid body with forces in a single plane (e.g., X-Y plane):

1. Translational equilibrium requires:

   • Sum of forces in x-direction = 0: $\sum F_x = 0$
   • Sum of forces in y-direction = 0: $\sum F_y = 0$

2. Rotational equilibrium requires:

   • Sum of torques about any axis perpendicular to the plane = 0: $\sum \tau_z = 0$

Note: When all forces lie in one plane, you only need to check torque about one axis perpendicular to that plane. If the resultant force is zero, then the total torque will be the same about any perpendicular axis.

The angular momentum of a particle depends critically on the chosen reference point:

• For a given particle moving with velocity v:

  • If reference point O lies directly on the line of motion: l = 0
  • If reference point O is at perpendicular distance r from the line of motion: l = mvr
  • If reference point O is at position requiring angle θ: l = mvr sin θ

• Key consequences:

  • Angular momentum is not an intrinsic property of the particle
  • Different observers using different reference points will calculate different angular momenta
  • The perpendicular distance from the reference point to the line of motion (r sin θ) is the critical factor
  • Angular momentum can be zero, positive, or negative depending on reference point location

Example: A car moving along a straight road has zero angular momentum relative to a point on the road, but nonzero angular momentum relative to a point away from the road.

Comparison:

• Solid circular plate: $I = \frac{1}{2}MR^2$
• Circular ring (thin): $I = MR^2$

The ring's moment of inertia is twice that of the plate despite having the same mass and outer radius because:

Physical reasoning:

1. Mass distribution: In the ring, all mass is concentrated at distance R from the axis, while in the plate, mass is distributed from 0 to R
2. Squared distance effect: Moment of inertia depends on $r^2$, so mass farther from the axis contributes disproportionately more
3. Weighted average: The plate's moment of inertia represents a weighted average of $r^2$ values from 0 to R
4. Parallel axis contribution: None of the mass in a ring needs the parallel axis theorem adjustment

Application: This principle is used in designing efficient flywheels, where placing mass at the periphery maximizes rotational inertia for a given amount of material.