Introduction
Rotating bodies possess kinetic energy due to motion of their mass elements. This rotational kinetic energy combines with translational kinetic energy for rolling objects. The work-energy theorem applies to rotations with torque doing work. Energy methods often simplify rotational problems.
Rotational Kinetic Energy
K_rot = 1/2Iω2. Analogous to K_trans = 1/2mv2 with I replacing m and ω replacing v. Depends on distribution of mass - same mass at larger radius gives larger I and more rotational KE. For fixed ω, object with larger I has more rotational energy.
Total Kinetic Energy
K_total = K_trans + K_rot = 1/2Mv2_CM + 1/2I_CMω2. For rolling without slipping: substitute v = Rω and use I about CM. Energy distributes between translation and rotation based on shape factor I/(MR2). Sphere: (2/5), disk: (1/2), ring: 1.
Work by Torque
dW = τdθ. Power P = τω. Work done by torque through angular displacement: W = ∫τdθ. For constant torque: W = τΔθ. Work-energy theorem: W_net = ΔK_rot. For rolling objects, both torque work and force work contribute to total energy change.
\nIntroduction
Rotating bodies possess kinetic energy due to motion of their mass elements. This rotational kinetic energy combines with translational kinetic energy for rolling objects. The work-energy theorem applies to rotations with torque doing work. Energy methods often simplify rotational problems.
Rotational Kinetic Energy
K_rot = 1/2Iω2. Analogous to K_trans = 1/2mv2 with I replacing m and ω replacing v. Depends on distribution of mass - same mass at larger radius gives larger I and more rotational KE. For fixed ω, object with larger I has more rotational energy.
Total Kinetic Energy
K_total = K_trans + K_rot = 1/2Mv2_CM + 1/2I_CMω2. For rolling without slipping: substitute v = Rω and use I about CM. Energy distributes between translation and rotation based on shape factor I/(MR2). Sphere: (2/5), disk: (1/2), ring: 1.
Work by Torque
dW = τdθ. Power P = τω. Work done by torque through angular displacement: W = ∫τdθ. For constant torque: W = τΔθ. Work-energy theorem: W_net = ΔK_rot. For rolling objects, both torque work and force work contribute to total energy change.
\nEnergy Conservation with Rotation
E = K_trans + K_rot + U = constant if only conservative forces act. For rolling objects on inclines: mgh = 1/2mv2 + 1/2Iω2. Using ω = v/R: mgh = 1/2mv2(1 + I/(MR2)). Solves for final velocity without knowing time or acceleration - powerful energy method.
Applications
Rolling race: object with smaller I/(MR2) accelerates faster (sphere beats cylinder down incline). Flywheels store energy as rotation. Yo-yo: potential converts to translational and rotational kinetic energy. Spinning skater: work done by internal forces changes kinetic energy distribution.
\nSolved Example: Objects Rolling Down Incline
A solid sphere (I = 2/5 MR2), solid cylinder (I = 1/2 MR2), and ring (I = MR2), all with same mass M and radius R, roll down an incline of height h = 1 m without slipping. Find final velocity of each and determine the winner. Solution: Energy conservation: mgh = 1/2mv2 + 1/2Iω2 = 1/2mv2 + 1/2I(v/R)2 = 1/2mv2(1 + I/(MR2)). v2 = 2gh/(1 + I/(MR2)). Sphere: I/(MR2) = 2/5 = 0.4, so v2 = 2×9.8×1/1.4 = 14, v = 3.74 m/s. Cylinder: I/(MR2) = 0.5, so v2 = 19.6/1.5 = 13.07, v = 3.62 m/s. Ring: I/(MR2) = 1, so v2 = 19.6/2 = 9.8, v = 3.13 m/s. Winner: sphere is fastest, then cylinder, then ring. Object with smaller rotational inertia has less energy tied up in rotation, more available for translation, so it moves faster. Time to reach bottom: sphere wins at 3.74 m/s. Free sliding object would reach v = √(2gh) = 4.43 m/s. All rolling objects are slower due to rotational energy requirement.
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