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Q6. For the rotational system subjected to an applied torque Mo sin 25t shown in Figure 5, (i) derive the equation of motion using energy method; (ii) find the steady-state response; and (iii) for what values of Mo the forced amplitude of angular displacement of the bar will be less than 3°. 1/4 k L-2 m = 0.8 kg k 1x 104 N/m L= 40 cm Mosin wot Slender bar of mass m

Q6. For the rotational system subjected to an applied torque Mo sin 25t shown in Figure 5, (i) derive the equation of motion using energy method; (ii) find the steady-state response; and (iii) for what values of Mo the forced amplitude of angular displacement of the bar will be less than 3°. 1/4 k L-2 m = 0.8 kg k 1x 104 N/m L= 40 cm Mosin wot Slender bar of mass m

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Q6. For the rotational system subjected to an applied torque Mo sin 25t shown in Figure 5, (i) derive the equation of motion using energy method; (ii) find the steady-state response; and (iii) for what values of Mo the forced amplitude of angular displacement of the bar will be less than 3°. 1/4 k L-2 m = 0.8 kg k 1x 104 N/m L= 40 cm Mosin wot Slender bar of mass m
Transcribed Image Text:Q6. For the rotational system subjected to an applied torque Mo sin 25t shown in Figure 5, (i) derive the equation of motion using energy method; (ii) find the steady-state response; and (iii) for what values of Mo the forced amplitude of angular displacement of the bar will be less than 3°. 1/4 k L-2 m = 0.8 kg k 1x 104 N/m L= 40 cm Mosin wot Slender bar of mass m
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