Dynamics / Sheet 07
Rotational motion
Formula reference for mechanical engineering interviews. Moment of inertia is about mass distribution, not mass. Take moments only about a fixed axis or about the centre of mass.
Scope: Rigid bodies, plane motionSections: 12Units: SI, rad and rad/sRevision: A
59 formulas
Angular kinematics
Angular velocity
ω = dθ / dt
rad/s; always convert from rpm first
Angular acceleration
α = dω/dt = ω dω/dθ
Second form when time is not given
Constant α
ω = ω0 + α t
Direct analogue of v = u + at
Constant α
θ = ω0 t + ½ α t²
θ in radians
Constant α
ω² = ω0² + 2 α θ
When time is not wanted
Speed conversion
ω = 2 π N / 60
N in rpm. Also N = 9.55 ω
Revolutions
θ = 2 π n
n in revolutions
Linear and angular link
Arc length
s = r θ
Only valid with θ in radians
Tangential velocity
v = r ω
Perpendicular to the radius
Tangential acceleration
at = r α
Present only when the speed is changing
Centripetal acceleration
ac = v² / r = r ω²
Always toward the centre, even at constant speed
Total acceleration
a = √(at² + ac²)
The two components are perpendicular
Instantaneous centre
v = ω × r from the IC
Point of zero velocity; simplifies linkage analysis
Mass moment of inertia
I = Σ m r² = ∫ r² dm
Distribution, not mass. Doubling the radius quadruples I.
| Body | Axis | I |
|---|---|---|
| Point mass | distance r | m r² |
| Thin rod, length L | through centre | M L² / 12 |
| Thin rod, length L | through one end | M L² / 3 |
| Solid disc or cylinder | own axis | ½ M R² |
| Solid disc | diameter | ¼ M R² |
| Thin ring or hoop | own axis | M R² |
| Thick cylinder | own axis | ½ M (Ro² + Ri²) |
| Solid sphere | diameter | ⅖ M R² |
| Thin spherical shell | diameter | ⅔ M R² |
| Rectangular plate a × b | normal, through centre | M (a² + b²) / 12 |
Parallel axis theorem
I = Icm + M d²
Only valid starting from the centroidal value
Radius of gyration
k = √(I / M)
The single radius that would give the same I
Composite bodies
I = Σ (Icm,i + mi di²)
Subtract for holes and cutouts
Kinetics
Rotational second law
Σ T = I α
Fixed axis, or about the centre of mass. Nothing else
General plane motion
ΣF = M acm, ΣMcm = Icm α
Two equations, always write both
Torque
T = F r sin θ = F × r
Only the perpendicular component counts
Power
P = T ω
Same relation as the gearbox sheet
Work
W = ∫ T dθ
Constant torque: W = T θ
Time to accelerate
t = J Δω / T
Constant torque from rest to speed
Energy, momentum, impulse
Rotational kinetic energy
KE = ½ I ω²
Analogue of ½mv²
Rolling body total KE
½ M v² + ½ Icm ω²
Translation plus rotation
Angular momentum
L = I ω
Vector; direction by the right hand rule
Momentum principle
Σ T = dL / dt
L is conserved when ΣT = 0
Angular impulse
∫ T dt = ΔL = I Δω
For impacts and short torque pulses
Conservation example
I1 ω1 = I2 ω2
Skater, reaction wheel, retracting robot arm
Rolling motion
Rolling without slipping
v = r ω, a = r α
The kinematic constraint
Acceleration down an incline
a = g sinθ / (1 + I/MR²)
Shape only. Mass and radius cancel
Ranking
sphere 0.71, cylinder 0.67, ring 0.50
Multiples of g sinθ
Friction required
μ ≥ tanθ · (I/MR²) / (1 + I/MR²)
Sphere needs only (2/7) tanθ
Kinetic energy split
KE = ½ M v² (1 + I/MR²)
I/MR² is the fraction diverted into spin
Static friction in rolling without slipping does no work, because the contact point is instantaneously at rest. Energy is conserved even though friction is essential to the motion.
Drivetrain inertia
Reflected through a ratio
Jref = Jload / i²
Divided by the SQUARE of the ratio
Leadscrew load
Jeq = m (L / 2π)²
L is the lead in metres per revolution
Belt or pinion load
Jeq = m r²
r is the pitch radius
Motor torque required
T = (Jm + Jref) αm + Tload/(i η) + Tf
Acceleration, load, friction. Note αm = i αload
Inertia ratio target
Jref / Jmotor = 1 to 10
Too high is sluggish, too low is an oversized motor
Balancing and vibration
Unbalance force
F = m e ω²
Scales with speed SQUARED
Static balance
Σ m r = 0
Single plane; resultant force zero
Dynamic balance
Σ m r = 0 and Σ m r l = 0
Two planes; force and couple both zero
Balance quality grade
G = e ω (mm/s)
G6.3 general machinery, G2.5 precision
Critical speed
ωc = √(k/m) = √(g/δ)
δ is the static deflection under self weight
Critical speed, practical
Nc ≈ 946 / √δ rpm
δ in mm. Fast mental estimate
Operating band
below 0.75 Nc or above 1.4 Nc
If you must pass through, do it quickly
Torsional natural frequency
ωn = √(kt / J)
kt = GJ/L from the torsion sheet
Flywheels and gyroscopes
Coefficient of fluctuation
Cs = (ωmax - ωmin) / ωmean
Speed variation the machine tolerates
Flywheel energy
ΔE = I ωmean² Cs
Equivalently ΔE = 2 E Cs
Flywheel sizing
I = ΔE / (ωmean² Cs)
Put the mass at the rim: I ∝ R²
Gyroscopic couple
C = I ωspin ωprecession
Acts perpendicular to both axes
Centrifugal force
F = m ω² r = m v² / r
Sizes rim stress and bearing loads
Rim hoop stress
σ = ρ v² = ρ ω² R²
Sets the burst speed of a flywheel
Worked chain: sizing an actuator
Step 1
build I about the joint axis
Composite bodies plus parallel axis for every offset mass
Step 2
α from the motion profile
From ω and the accel time, or from ω² = 2αθ
Step 3
T = I α + Tgravity + Tfriction
Gravity term peaks with the arm horizontal
Step 4
reflect through the ratio, J/i²
Check the inertia ratio lands in 1 to 10
Step 5
check peak and RMS torque
Peak sizes the drive, RMS sizes the thermals
Step 6
check critical and torsional speeds
Stay clear of the operating band
Numbers worth memorizing
| Quantity | Value | Comment |
|---|---|---|
| rpm to rad/s | ω = N / 9.55 | 3000 rpm = 314 rad/s |
| Rod about end vs centre | 4× | ML²/3 against ML²/12 |
| Falling rod tip acceleration | 1.5 g | uniform rod released horizontal |
| Rolling ranking | 0.71 / 0.67 / 0.50 | sphere / cylinder / ring, times g sinθ |
| Balance grade, general | G6.3 | G2.5 for machine tool spindles |
| Critical speed estimate | 946 / √δ rpm | δ in mm |
| Inertia ratio target | 1 to 10 | reflected load to motor |
Concepts that decide the interview
- Moment of inertia is about distribution, not mass. Doubling the radius at constant mass quadruples I. That is why flywheels put mass at the rim and why robot arms put motors at the base.
- Take moments only about a fixed axis or about the centre of mass. Using an arbitrary accelerating point is the most common error in rigid body dynamics.
- Rolling acceleration depends on shape alone, because mass and radius cancel. The term I/MR² is the fraction of energy diverted into spin instead of translation.
- Static friction in rolling without slipping does no work, since the contact point is instantaneously at rest.
- The tip of a uniform rod released from horizontal accelerates at 1.5 g, faster than free fall, because the pivot supplies an upward force while the rod rotates about it.
- Reflected inertia divides by the ratio squared, which is what makes a geared joint controllable and direct drive hard.
- Unbalance force scales with ω², so a rotor that is fine at 1000 rpm can destroy bearings at 4000 rpm with no change in manufacture. That is why balance grades are specified as e·ω.
- Critical speed is a stiffness and mass problem, not a strength problem, and ωc = √(g/δ) makes it a fast mental estimate from static deflection alone.
- Angular momentum conservation is a design tool: reaction wheels, control moment gyros, and the gyroscopic couple that loads bearings whenever a machine carrying a fast rotor turns.
Sheet 07 of 07 / Rev ARigid bodies, plane motion, radians throughout