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A flywheel of mass 100 kg and radius 1 m is rotating at the rate of 420 rev/min. Find the constant retarding torque to stop the wheel in 14 revolutions, the mass is concentrated at the rim. M.I. of the flywheel about its axis of rotation I = mr^(2).

Answer» <html><body><p></p>Solution :Mass of flywheel = m = <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> kg <br/> Radius= <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> = 1m <br/> `I = mr^(2) = 100 xx 1^(2) = 100 kg m^(2)` <br/> Initial angular velocity `=omega_(0) = 2pir = 2 xx 3.14 xx 420/60 = 43.96 "rad"//s^(2)` <br/> Final angular velocity `=omega =0` <br/> Angular displacement in 14 <a href="https://interviewquestions.tuteehub.com/tag/revolution-623158" style="font-weight:bold;" target="_blank" title="Click to know more about REVOLUTION">REVOLUTION</a> = `14 xx <a href="https://interviewquestions.tuteehub.com/tag/2pi-1838601" style="font-weight:bold;" target="_blank" title="Click to know more about 2PI">2PI</a> = 28 pi` radian. <br/> `alpha = (omega^(2) - omega_(0)^(2))/(<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>) = (0-43.96^(2))/(2 xx 28 pi) =-10.99 "rad"//s^(2)`<br/> Torque required to stop the flywheel `=tau = Ialpha = 100 xx 10.99 = 1099` Nm</body></html>


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