• # question_answer Two rotors A and B are connected to the two ends of a shaft of uniform diameter. The mass moment of inertia of rotor A about the axis of the shaft is four times that of B. If the length of the shaft is 1 m and C is the position of node for torsional vibrations, then what is the length of AC? A) 1/5 m               B) 4/5 mC) 1/25 m             D) 16/25 m

Solution :

$AC=\left( \frac{{{I}_{B}}}{{{I}_{A}}+{{I}_{B}}} \right)\,l=\left( \frac{{{I}_{B}}}{4{{I}_{B}}+{{I}_{B}}} \right)\,l$$=\frac{l}{5}=\frac{1}{5}\,m$

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