Rotational Mechanics

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.

• 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.

The angular position (θ) in rotational motion:

• Represents the angle through which a rigid body has rotated from its initial position
• Is measured between a reference line and the current position line
• Is the same for all particles in a rigid body during rotation
• Can be related to linear displacement (s) along circular path by s = rθ, where r is the perpendicular distance from the particle to the axis of rotation

When a particle moves from position P₀ to P during rotation, the entire rigid body rotates through angle θ about the axis of rotation.

The torque of a force F about an axis AB will be zero under these conditions:

1. When F is parallel to axis AB

   • Since r × F is perpendicular to F, and F is parallel to AB, the torque vector has no component along AB

2. When F intersects the axis AB

   • If F passes through any point on the axis, taking that intersection point as origin gives r = 0
   • This makes the cross product r × F = 0

3. When F lies in a plane containing the axis AB

   • The torque vector will be perpendicular to this plane
   • Thus it will be perpendicular to the axis, giving zero component along the axis

4. When the force and axis are in the same plane and F is directed toward the axis

   • The perpendicular distance (lever arm) becomes zero
   • Example: A radial force in a rotating system produces no torque about the center of rotation

These conditions are crucial for analyzing mechanical equilibrium and designing systems where certain forces should not contribute to rotation.

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.

When a cyclist takes a horizontal circular turn:

• The cyclist must lean inward to maintain rotational equilibrium
• The angle of inclination (θ) is determined by: tan(θ) = v²/rg

Where:

• v = velocity of the cyclist
• r = radius of the circular path
• g = acceleration due to gravity

This happens because:

1. The centripetal force needed (Mv²/r) must be provided by the horizontal component of the normal force
2. The vertical component of the normal force must balance the weight (Mg)
3. Without leaning, the cyclist would experience an unbalanced torque and fall outward

Example: A cyclist moving at 10 m/s on a turn with radius 25 m would need to lean at an angle where tan(θ) = 10²/(25×9.8) = 0.408, or approximately 22.2°

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 particle of mass m moving in a circular path of radius r with tangential velocity v:

Angular momentum = mvr

Key relationships:

• The direction of angular momentum is perpendicular to both r and v (follows right-hand rule)
• If the particle moves with constant speed, its angular momentum magnitude remains constant
• The angular momentum can be expressed as L = mr²ω, where ω is the angular velocity

Example: A satellite orbiting Earth maintains constant angular momentum in the absence of external torques. If the orbit radius decreases, the satellite's velocity must increase to conserve angular momentum.

Angular impulse equals the change in angular momentum:

$J = \int_{t_1}^{t_2} \Gamma \, dt = L_2 - L_1$

Where:

• $J$ is the angular impulse
• $\Gamma$ is the applied torque
• $L_1$ and $L_2$ are the initial and final angular momenta

This is analogous to linear impulse-momentum relationship but applied to rotational systems.

To calculate the moment of inertia of a uniform solid sphere about a diameter:

1. Consider the sphere as composed of infinitesimally thin concentric spherical shells

2. For a shell of radius x and thickness dx:

   • Mass of shell = (3M/R³)x²dx (where M is total mass, R is sphere radius)
   • Moment of inertia of this shell = (2/3)x² × shell mass
   • dI = (2/3)x² × (3M/R³)x²dx = (2M/R³)x⁴dx

3. Integrate over all shells from center to surface: I = ∫₀ᴿ (2M/R³)x⁴dx = (2M/R³)(R⁵/5) = (2/5)MR²

This result (2/5)MR² applies to any diameter due to the sphere's symmetry.