A fly wheel of mass 10 kg and diameter 0.4 m rotating at 120 rpm has its speed increased to 720 rpm in 8 seconds. Find the torque acting on fly wheel.

Explanation

Given, m= 10 kg, D= 0.4m,` R= (0.4)/(2) = 0.2 m`
` omega_0 = 120 "rpm" = (120)/(60) = 2 rps`
` omega = 720 " rpm" = 720/60 = 7 rps`
` t = 8s , tau = ? `
moment of inertatia `I = ? alpha= ? `
w.k.t ` omega = omega_0 + alpha t `
` therefore alpha = (omega - omega_0)/(t) = (7-2)/(8) = 5/8 rps^2`
i.e., ` alpha = 5/8 xx 2 pi rad s^(-2)`
` alpha = (5pi)/(4) rad s^(-2)`
Moment of inertia ` I = (MR^2)/(2) = (10 xx 0.2 xx 0.2)/(2)`
i.e., `I = 5 xx 0.4 = 0.20 kgm^2`
` tau = Ialpha = (0.2 xx 5pi)/(4)`
` tau = 0.05 xx 5 xx 3.14 = 0.7855`
Torque acting on fly wheel = 0.785 Nm