A disk having mass of 50 kg and diameter 0-2 m is mounted on a shaft- If due mounting error it makes angle 1- with plane normal to shaft axis- Determine gyroscopic couple acting on the bearing assuming rotation speed of shaft as 125 rad-sec-

Gyroscopic couple experienced by bearing due to misalignment of disk is given by G= m 8×ω^2× r^2×sin(2θ) Substitute respective values and we get G = 34 N m ...

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Standard XII

Physics

Friction

A disk having...

Question

$A disk having mass of 50 kg and diameter 0.2 m is mounted on a shaft. If due mounting error it makes angle 1^{\circ} with plane normal to shaft axis. Determine gyroscopic couple acting on the bearing assuming rotation speed of shaft as 125 rad/sec?$

34 N-m

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68 N-m

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102 N-m

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136 N-m

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Solution

The correct option is A 34 N-m
Gyroscopic couple experienced by bearing due to misalignment of disk is given by

$G = \frac{m}{8} \times ω^{2} \times r^{2} \times sin (2 θ)$

Substitute respective values and we get G = 34 N-m

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A disk having mass of 50 kg and diameter 0.2 m is mounted on a shaft. If due mounting error it makes angle 1∘ with plane normal to shaft axis. Determine gyroscopic couple acting on the bearing assuming rotation speed of shaft as 125 rad/sec?

The correct option is A 34 N-mGyroscopic couple experienced by bearing due to misalignment of disk is given by G=m8×ω2×r2×sin(2θ) Substitute respective values and we get G = 34 N-m

$A disk having mass of 50 kg and diameter 0.2 m is mounted on a shaft. If due mounting error it makes angle 1^{\circ} with plane normal to shaft axis. Determine gyroscopic couple acting on the bearing assuming rotation speed of shaft as 125 rad/sec?$

The correct option is A 34 N-m
Gyroscopic couple experienced by bearing due to misalignment of disk is given by

$G = \frac{m}{8} \times ω^{2} \times r^{2} \times sin (2 θ)$

Substitute respective values and we get G = 34 N-m