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A fly wheel of mass 12.5kg and diameter 0.36 m rotating at 90rpm has its speed increased to 720rpm in 8s. Find the torque applied to flywheel |
| Answer» EAR,◆ Answer-τ = 3.34 Nm◆ Explanation-# Given-m = 12.5 kgr = 0.36/2 = 0.18 mf1 = 90 rpm = 1.5 rpsf2 = 720 rpm = 12 rpst = 8 s# Solution-Rotational acceleration is-α = (ω2-ω1) / tα = 2π(f2-f1) / tα = 2π(12-1.5) / 8α = 8.247 rad/s^2Moment of INERTIA is -I = mr^2I = 12.5 × 0.18^2I = 0.405 kgm^2Rotational TORQUE is calculated by-τ = Iατ = 0.405 × 8.247τ = 3.34 NmTherefore, torque required is 3.34 Nm.Hope this helps you... | |