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A uniform horizontal disc fixed at its centre to an elastic vertical rod performs forced torsional oscillations dur to the moment of forces N_(z)=N_(m)cos omegat. The oscialltions obey the law varphi=varphi_(m) cos ( omega t- alpha). (a) the work performed by friction forces acting on the disc during one oscillation period ,(b) the quality factor of the given oscillator if th emoment of inertia of the disc relative to the axis is equal to I. |
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Answer» Solution :The equation of the disc is `ddot(varphi)+ 2 beta dot (varphi)+ omega_(0)^(2) varphi=(N_(m) cos omegat)/( I)` Then as before `varphi=varphi_(m) cos ( omegat- ALPHA)` where `varphi(m)=(N_(m))/(I[( omega_(0)^(2)-OMEGA^(2))^(2)+ 4 beta^(2)omega^(2)] ^(1//2)), tan alpha=(2 beta omega)/( omega_(0)^(2)- omega^(2))` `(a)` Work performed by frictional forces `=-int N_(R) d varphi` where`N_(r)=-2 I beta dot(varphi)=- int_(0)^(T) 2 betaI dot(varphi^(2))dt=-2 pi beta omega I varphi_(m)^(2)` `=- pi I varphi_(m)^(2)[( omega_(0)^(2)- omega^(2))^(2)+4 beta^(2)omega^(2)]^(1//2)SIN alpha=- pi N_(m) varphi_(m) sin alpha` `(b)` The quality factor `Q=(pi)/(lambda)=(pi)/(betaT)=(sqrt(omega_(0)^(2)-beta^(2)))/( 2 beta)=( omegasqrt(omega_(0)^(2)-beta^(2)))/( ( omega_(0)^(2)- omega^(2) ) tan alpha) =(1)/(tan alpha){( 4 omega^(2) omega_(0)^(2))/((omega_(0)^(2)-omega^(2))^(2))-(4 beta^(2)omega^(2))/((omega_(0)^(2)-omega^(2))^(2))}` `=(1)/(2 tan alpha) { ( 4 omega^(2) omega_(0)^(2) I^(2) varphi_(m)^(2))/( N_(m)^(2)cos^(2) alpha)- tan ^(2) alpha}`since`omega_(0)^(2)= omega^(2)+(N)/( Ivarphi_(m))cos alpha` `=(1)/(2 sin alpha){(4 omega^(2) omega_(0)^(2) I^(2) varphi_(m)^(2))/( N_(m)^(2)) - sin ^(2) alpha}^(1//2)` `=(1)/(2 sin alpha){ ( 4 omega^(2) I^(2) varphi_(m)^(2))/( N_(m)^(2))(omega^(2)+(N_(m) cos alpha)/( I varphi_(m)))+ 1- cos^(2) alpha}^(1//2)` `=(1)/( 2sin alpha){( 4 I^(2) varphi_(m)^(2))/( N_(m)^(2))omega^(4)+ ( 4Ivarphi_(m))/( N_(m))omega ^(2) cos alpha+ cos ^(2) alpha-1}^(1//2)=(1)/( 2sin alpha ){((2Ivarphi_(m)omega^(2))/( N_(m))+ cos alpha)^(2)-1}^(1//2)` |
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