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Introduction Rotational Kinematics Linear and Angular Relationship Rotational Kinematics Equations Torque Moment of Inertia Newton's Second Law for Rotation Rotational Kinetic Energy Work Done by Torque Angular Momentum Conservation of Angular Momentum Equilibrium of Rigid Bodies Rolling MotionUntil now, we've treated objects as point particles where all mass concentrates at a single point. This simplification works well for translational motion, but real objects have size and shape. They can rotate, spin, roll, and pivot, not just slide.
Rigid body: An idealized extended object in which the distance between any two points remains constant regardless of external forces. While no real object is perfectly rigid (all materials deform slightly under stress), many objects approximate rigid body behavior well enough for practical analysis: wheels, doors, rods, gears, planets.
Rigid body mechanics extends Newton's laws to rotational motion.
Angular displacement (θ): The angle through which a point, line, or body has rotated about a specified axis.
SI unit: radian (rad), where 2π rad = 360° = 1 complete revolution
Angular velocity (ω): The rate of change of angular displacement.
ω = dθ/dt (instantaneous) | ω_avg = Δθ/Δt (average)
SI unit: rad/s (radians per second)
Angular acceleration (α): The rate of change of angular velocity.
α = dω/dt (instantaneous) | α_avg = Δω/Δt (average)
SI unit: rad/s² (radians per second squared)
s = rθ (arc length)
v = rω (tangential speed)
a_t = rα (tangential acceleration)
a_c = v²/r = rω² (centripetal acceleration)
Example: A wheel of radius 0.3 m rotates at 5 rad/s. Find the linear speed of a point on its rim.
v = rω = (0.3)(5) = 1.5 m/s
When angular acceleration is constant, rotational kinematics equations directly parallel linear equations:
| Linear | Rotational |
|---|---|
| v = u + at | ω = ω₀ + αt |
| s = ut + ½at² | θ = ω₀t + ½αt² |
| v² = u² + 2as | ω² = ω₀² + 2αθ |
| s = ½(u + v)t | θ = ½(ω₀ + ω)t |
τ = r × F (vector product)
Magnitude: τ = rF sin θ
r = lever arm (perpendicular distance to axis)
θ = angle between r and F
Torque is the rotational equivalent of force. It's the tendency of a force to cause rotation about an axis.
SI unit: N·m (newton-meter)
Sign convention:
Example: You push a door with 20 N perpendicular to its surface at a point 0.8 m from the hinges. Calculate the torque.
τ = rF sin 90° = (0.8)(20)(1) = 16 N·m
For a point mass: I = mr²
For extended objects: I = Σmᵢrᵢ² or I = ∫r²dm
Moment of inertia (rotational inertia) is the rotational analog of mass. It measures an object's resistance to angular acceleration about a particular axis.
SI unit: kg·m²
The moment of inertia depends on:
| Object | Axis | Moment of Inertia |
|---|---|---|
| Thin rod (length L) | Through center, perpendicular | I = (1/12)ML² |
| Thin rod (length L) | Through end, perpendicular | I = (1/3)ML² |
| Solid sphere (radius R) | Through center | I = (2/5)MR² |
| Hollow sphere (radius R) | Through center | I = (2/3)MR² |
| Solid cylinder (radius R) | Through central axis | I = (1/2)MR² |
| Thin hoop (radius R) | Through center, perpendicular | I = MR² |
Parallel Axis Theorem: I = I_cm + Md²
I_cm = moment about center of mass, d = distance between parallel axes
τ_net = Iα
This is the rotational analog of F_net = ma.
Example: A wheel with I = 0.5 kg·m² experiences a net torque of 2 N·m. Find its angular acceleration.
α = τ_net/I = 2/0.5 = 4 rad/s²
KE_rot = ½Iω²
For an object undergoing both translation and rotation (like a rolling wheel):
KE_total = ½mv²_cm + ½I_cm ω²
Example: A solid cylinder (M = 2 kg, R = 0.1 m) rolls without slipping at v = 3 m/s. Calculate total kinetic energy.
I = ½MR² = ½(2)(0.1)² = 0.01 kg·m²
v = Rω → ω = 3/0.1 = 30 rad/s
KE_trans = ½(2)(3)² = 9 J
KE_rot = ½(0.01)(30)² = 4.5 J
KE_total = 13.5 J
W = τθ (for constant torque)
P = τω (power delivered by torque)
θ must be in radians.
L = Iω (for rigid body rotating about fixed axis)
L = r × p = mvr sin θ (for a point mass)
Angular momentum is the rotational analog of linear momentum.
SI unit: kg·m²/s or J·s
If τ_ext = 0, then L = constant
ΣL_initial = ΣL_final
Example: Figure skater spin
Skater spins with arms extended (large I, modest ω). Pulling arms in decreases I. To conserve L = Iω, ω must increase — the skater spins faster.
If I_f = ½I_i, then ω_f = 2ω_i (angular velocity doubles).
For a rigid body in complete equilibrium:
Example: A uniform beam (length 5 m, mass 100 kg) is supported at its center. A 50 kg person stands 1.5 m from the left end. Where should a 60 kg person stand to balance?
Taking torques about the support point (center):
τ_cw = (50)(9.81)(1.0) = 490.5 N·m
τ_ccw = (60)(9.81)(x) = 490.5 → x = 0.833 m to the right of center
Rolling without slipping: The contact point between the rolling object and surface is instantaneously at rest.
v_cm = Rω
For an object rolling down an incline from rest through vertical height h:
mgh = ½mv²_cm + ½I_cm ω²
mgh = ½mv²_cm(1 + I_cm/(MR²))
Objects with smaller I/(MR²) accelerate faster down inclines.