Source: ENGR 216 Lectures, Lab 5
Tags: rotational motion, angular velocity, angular acceleration, moment of inertia, angular momentum, centre of mass, parallel axis theorem, torque, rotational kinetic energy, ENGR 216, PHYS 206
Rotational motion mirrors linear motion: torque replaces force, moment of inertia replaces mass, and angular acceleration replaces linear acceleration. The parallel axis theorem lets you find the moment of inertia about any axis once you know it about the centre of mass. Angular momentum is conserved when no net external torque acts, which explains why spinning objects speed up when they pull mass inward.
Angular displacement (θ)
The angle through which an object rotates, measured in radians (rad). One full revolution = 2π rad.
Angular velocity (ω)
The rate of change of angular displacement: ω = dθ/dt, measured in rad/s. Positive is counterclockwise by convention.
Angular acceleration (α)
The rate of change of angular velocity: α = dω/dt, measured in rad/s².
Moment of inertia (I)
The rotational analogue of mass. It measures resistance to angular acceleration. Depends on both mass and how that mass is distributed relative to the axis of rotation: I = Σm_i r_i² (discrete) or I = ∫r² dm (continuous). Units: kg·m².
Torque (τ)
The rotational analogue of force: τ = r × F, or τ = rF sin θ. Newton's second law for rotation is Στ = Iα.
Angular momentum (L)
The rotational analogue of linear momentum: L = Iω (for rotation about a fixed axis). Units: kg·m²/s.
Conservation of angular momentum
When no net external torque acts on a system, total angular momentum is constant: L_initial = L_final, or I₁ω₁ = I₂ω₂.
Centre of mass (COM)
The point where the total mass of a system can be considered to be concentrated. For a collection of particles: x_COM = Σ(m_i x_i) / Σm_i, and similarly for y and z.
Parallel axis theorem
Relates the moment of inertia about any axis to the moment of inertia about a parallel axis through the centre of mass: I = I_COM + Md², where d is the distance between the two parallel axes and M is the total mass.
Rotational kinetic energy
The kinetic energy due to rotation: KE_rot = ½Iω². For a rolling object, total KE = ½mv² + ½Iω².
When angular acceleration is constant, the rotational kinematic equations are identical in structure to their linear counterparts:
ω = ω₀ + αt
θ = θ₀ + ω₀t + ½αt²
ω² = ω₀² + 2α(θ − θ₀)
θ = θ₀ + ½(ω₀ + ω)t
The correspondence is: x → θ, v → ω, a → α.
For a point on a rotating object at distance r from the axis:
Arc length: s = rθ
Tangential velocity: v_t = rω
Tangential acceleration: a_t = rα
Centripetal acceleration: a_c = v²/r = rω²
These are standard results you should know or be able to look up quickly:
Solid cylinder (or disk) about its central axis: I = ½MR²
Hollow cylinder (thin ring) about its central axis: I = MR²
Solid sphere about any diameter: I = ⅖MR²
Hollow sphere about any diameter: I = ⅔MR²
Thin rod about its centre: I = (1/12)ML²
Thin rod about one end: I = (1/3)ML²
Point mass at distance r: I = mr²
If you know I_COM for a shape and need I about a different parallel axis:
I = I_COM + Md²
This is commonly needed when combining shapes. Find I_COM for each piece about its own centre of mass, then shift each to the common axis using the parallel axis theorem, and add them together.
Example: a thin rod of mass M and length L, pivoted about one end. I_COM = (1/12)ML², and d = L/2 (distance from centre to end). So I_end = (1/12)ML² + M(L/2)² = (1/12)ML² + (1/4)ML² = (1/3)ML².
Στ = Iα
This is the rotational analogue of ΣF = ma. Apply it exactly the same way: draw the free body diagram, identify all torques about the chosen axis, sum them (with sign convention), set equal to Iα, and solve.
When no net external torque acts:
I₁ω₁ = I₂ω₂
The classic example is an ice skater pulling their arms in. Reducing r decreases I, so ω must increase to keep L constant. This principle is directly relevant to Lab 5 (Rotational Motion).
An object that rolls without slipping satisfies v_COM = Rω. The total kinetic energy is translational plus rotational:
KE_total = ½mv² + ½Iω²
For a solid disk rolling down a ramp, using energy conservation:
mgh = ½mv² + ½(½mR²)(v/R)² = ½mv² + ¼mv² = ¾mv²
So v = √(4gh/3), which is slower than a sliding object (v = √(2gh)) because energy goes into rotation.
Rotational kinematics (constant α): ω = ω₀ + αt θ = θ₀ + ω₀t + ½αt² ω² = ω₀² + 2α(θ − θ₀)
Torque: τ = rF sin θ , Στ = Iα
Moment of inertia: I = Σm_i r_i² , I = I_COM + Md²
Angular momentum: L = Iω
Rotational kinetic energy: KE_rot = ½Iω²
Rolling constraint: v = Rω (no slipping)
⚠️ Moment of inertia depends on the axis of rotation. The same object has different I values about different axes. Always specify the axis.
⚠️ The parallel axis theorem only works to shift from the centre-of-mass axis to another parallel axis (or vice versa). You cannot use it to shift between two arbitrary parallel axes directly; go through the COM axis as an intermediate step.
⚠️ For rolling without slipping, v = Rω. If the problem says "rolls without slipping," this constraint is essential. If it says "slides," there is no such constraint.
⚠️ Conservation of angular momentum questions are common: a turntable problem, a skater, or a collision where an object sticks to a rotating platform.
⚠️ Lab 5 directly tests rotational motion concepts. You will measure angular quantities and verify Στ = Iα or conservation of L experimentally.
Q: A solid disk of mass 4 kg and radius 0.3 m rotates about its central axis. What is its moment of inertia?
A: I = ½MR² = ½(4)(0.3²) = 0.18 kg·m².
Q: A figure skater spins at 2 rev/s with arms extended (I = 5 kg·m²). She pulls her arms in, reducing I to 2 kg·m². What is her new spin rate?
A: Conservation of angular momentum: I₁ω₁ = I₂ω₂. So ω₂ = (5 × 2) / 2 = 5 rev/s.
Q: What is the parallel axis theorem and when do you use it?
A: I = I_COM + Md². Use it when you need the moment of inertia about an axis that is parallel to, but offset from, the centre-of-mass axis.
Q: A torque of 12 N·m is applied to a wheel with I = 3 kg·m². What is the angular acceleration?
A: α = τ/I = 12/3 = 4 rad/s².
Q: Why does a solid sphere roll down a ramp faster than a hollow sphere of the same mass and radius?
A: The solid sphere has a smaller moment of inertia (⅖MR² vs. ⅔MR²), so less energy goes into rotation and more into translation, giving it a higher speed at the bottom.
rotational motion, angular velocity, angular acceleration, moment of inertia, torque, angular momentum, conservation of angular momentum, parallel axis theorem, centre of mass, center of mass, rotational kinetic energy, rolling without slipping, solid disk, hollow cylinder, thin rod, Newton's second law for rotation, Lab 5, ENGR 216, PHYS 206, mechanics, Texas A&M