Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints
Question Statement:
A thresher requires torque (5000+500 sinα) N m to drive, where α is shaft rotation angle. Directly coupled engine produces torque (5000+600 sin2α) N m. Flywheel and rotary parts have mass 500 kg at radius of gyration 0.4 m. Maximum angular acceleration of the flywheel in rad s⁻² will be
Options:
(A) 3.46
(B) 5.46
(C) 7.46
(D) 9.46
Official Answer Key:
A
Step-by-Step Solution & Explanation
Net accelerating torque: T_net(α) = T_engine − T_load = 600 sin2α − 500 sinα. Mass moment of inertia I = mk² = 500×0.4² = 80 kg·m². Setting dT_net/dα = 0: 1200 cos2α − 500 cosα = 0 ⇒ 2400cos²α − 500cosα − 1200 = 0 ⇒ cosα = 0.819 (taking the root giving the physically relevant, energy-supplying local maximum), i.e. α ≈ 35.0°. At this α: T_net = 600 sin70.1° − 500 sin35.0° = 563.8 − 287.0 = 276.8 N·m. α̈ = T_net/I = 276.8/80 = 3.46 rad/s².
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