Determine the acceleration of each mass and the angular acceleration of the pulley (no friction on the point like pivot). The following data is assumed to be given. Moment of inertia of pulley relative to the center of mass=2kgm2, M1=M2=0.5kg, R2=2R1=0.5m
T=i*alpha ,. M1g-T1=M1a1, T2-m2g=m2a2, ,alpha=a1/R1,alpha=a2/R2
T2R2-T1R1=i*alpha,T1+T2=2×0.5×9.81=9.81
M1g-T1=M1alphaR1,T2-m2g=m2alphaR2
M2gR2+M2alphaR2^2-M1gR1+M1R1^2 alpha=2alpha
(M2gR2-M1gR1)/(2-M1R1^2-M2R2^2)
=(9.81×0.5×0.5-9.81×0.25×0.5)/(2-0.5×0.25-0.5×0.0625)=
=0.67rad/s2
at2=alpha×R=0.67×0.5=0.33m/s2
at1=0.165m/s2
Determine the acceleration of each mass and the angular acceleration of the pulley (no friction on...
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