A shaft, supported on two bearings at its ends, carries two flywheels 'L' apart. Mass moment of inertia of the two flywheels are $${{l}_{a}}$$ and $${{l}_{b}}$$ I being the polar moment of inertia of cross-sectional area of inertia of shaft. Distance $${{l}_{a}}$$ of the node of torsional vibration of the shaft from the flywheel la is $${{l}_{a}}$$ given by:

Question

TuteeHUB · Accepted Answer

$${{l}_{a}}=\frac{L{{l}_{a}}}{{{I}_{a}}+{{I}_{b}}}$$

TuteeHUB · Answer

$${{l}_{a}}=\frac{L{{l}_{b}}}{{{I}_{a}}+{{I}_{b}}}$$

TuteeHUB · Answer

$${{l}_{a}}=\frac{L{{l}_{b}}}{{{I}_{a}}+{{I}_{b}}-1}$$

TuteeHUB · Answer

$${{l}_{a}}=\frac{L{{l}_{a}}}{{{I}_{a}}+{{I}_{b}}-1}$$

1.	A shaft, supported on two bearings at its ends, carries two flywheels 'L' apart. Mass moment of inertia of the two flywheels are \[{{l}_{a}}\] and \[{{l}_{b}}\] I being the polar moment of inertia of cross-sectional area of inertia of shaft. Distance \[{{l}_{a}}\] of the node of torsional vibration of the shaft from the flywheel la is \[{{l}_{a}}\] given by:
A.	\[{{l}_{a}}=\frac{L{{l}_{b}}}{{{I}_{a}}+{{I}_{b}}}\]
B.	\[{{l}_{a}}=\frac{L{{l}_{a}}}{{{I}_{a}}+{{I}_{b}}}\]
C.	\[{{l}_{a}}=\frac{L{{l}_{b}}}{{{I}_{a}}+{{I}_{b}}-1}\]
D.	\[{{l}_{a}}=\frac{L{{l}_{a}}}{{{I}_{a}}+{{I}_{b}}-1}\]
Answer» B. \[{{l}_{a}}=\frac{L{{l}_{a}}}{{{I}_{a}}+{{I}_{b}}}\]

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