Introduction
Torque produces angular acceleration. For fixed-axis rotation, τ = Iα is the rotational analog of F = ma. Understanding this relationship is fundamental to analyzing rotating rigid bodies, from simple wheels to complex machinery and celestial objects.
Rotational Dynamics Equation
For fixed axis: Στ = Iα where τ is torque about the axis, I is moment of inertia about same axis, α is angular acceleration. All quantities measured about same fixed axis. This is Newton's Second Law for rotation.
Torque Calculation
τ = r × F. Magnitude: τ = rFsinθ = r_⊥F = rF_⊥ where r_⊥ is moment arm. Direction by right-hand rule. For multiple forces: vector sum of individual torques. Same force produces different torque depending on application point and angle.
Rotational Kinematics
Analogous to linear: ω = ω0 + αt, θ = ω0t + 1/2αt2, ω2 = ω02 + 2αθ. Valid for constant angular acceleration. Connects angular displacement, velocity, and acceleration. Useful for problems where torque is constant (giving constant α).
\nIntroduction
Torque produces angular acceleration. For fixed-axis rotation, τ = Iα is the rotational analog of F = ma. Understanding this relationship is fundamental to analyzing rotating rigid bodies, from simple wheels to complex machinery and celestial objects.
Rotational Dynamics Equation
For fixed axis: Στ = Iα where τ is torque about the axis, I is moment of inertia about same axis, α is angular acceleration. All quantities measured about same fixed axis. This is Newton's Second Law for rotation.
Torque Calculation
τ = r × F. Magnitude: τ = rFsinθ = r_⊥F = rF_⊥ where r_⊥ is moment arm. Direction by right-hand rule. For multiple forces: vector sum of individual torques. Same force produces different torque depending on application point and angle.
Rotational Kinematics
Analogous to linear: ω = ω0 + αt, θ = ω0t + 1/2αt2, ω2 = ω02 + 2αθ. Valid for constant angular acceleration. Connects angular displacement, velocity, and acceleration. Useful for problems where torque is constant (giving constant α).
\nRolling Without Slipping
Condition: v_CM = Rω and a_CM = Rα. Kinetic energy: K = 1/2mv2 + 1/2Iω2 = 1/2mv2(1 + I/(mR2)). Factor in parentheses > 1 indicates rotational contribution. For solid sphere rolling: I = (2/5)mR2, so K = (7/10)mv2. Rolling objects accelerate slower than sliding due to energy going into rotation.
Atwood Machine with Massive Pulley
Equations: m1g - T1 = m1a, T2 - m2g = m2a, (T1 - T2)R = Iα = I(a/R). Solve for acceleration a = (m1-m2)g/(m1+m2+I/R2). Effective mass increased by rotational inertia term I/R2. System accelerates slower than with massless pulley.
\nSolved Example: Yo-Yo Descent
A yo-yo of mass m = 0.2 kg has outer radius R = 3 cm and inner axle radius r = 0.5 cm. Approximate as solid disk (I = 1/2mR2). Released from rest, find: (a) Acceleration, (b) Tension in string, (c) Velocity after falling 1 m. Solution: Equations: mg - T = ma (linear), Tr = Iα = (1/2mR2)(a/r). From torque: T = (1/2mR2)(a/r)/r = 1/2m(R/r)2a. Substitute: mg - 1/2m(R/r)2a = ma → g = a[1 + 1/2(R/r)2]. (R/r) = 6, so (R/r)2 = 36. g = a[1 + 18] = 19a → a = g/19 = 9.8/19 = 0.516 m/s2. Much less than g due to rotation. (b) T = 1/2m(R/r)2a = 1/2×0.2×36×0.516 = 1.86 N. Compare to mg = 1.96 N. Tension almost equals weight - most of gravitational force goes into rotation, little into linear acceleration. (c) v2 = 2ah = 2×0.516×1 = 1.032 → v = 1.02 m/s. Free fall would give v = √(2×9.8×1) = 4.43 m/s. Yo-yo is 4× slower due to rotational inertia.
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