A table rotates about a vertical axis passing | vedclass

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Question

(C) Since the axis of the table is vertical,its angular velocity $\vec{\omega}_T$ is in the vertical direction.The axis of the flywheel is horizontal,

Vedclass · Accepted Answer

$20\sqrt{5}\ rad/s$

Vedclass · Answer

(C) Since the axis of the table is vertical,its angular velocity $\vec{\omega}_T$ is in the vertical direction.The axis of the flywheel is horizontal,so its angular velocity $\vec{\omega}_F$ is in the horizontal direction.The resultant angular velocity $\vec{\omega}_R$ is the vector sum of these two perpendicular angular velocities:$\vec{\omega}_R = \vec{\omega}_F + \vec{\omega}_T$Since $\vec{\omega}_F$ and $\vec{\omega}_T$ are perpendicular,the magnitude of the resultant angular velocity is:$|\vec{\omega}_R| = \sqrt{\omega_F^2 + \omega_T^2}$$|\vec{\omega}_R| = \sqrt{40^2 + 20^2} = \sqrt{1600 + 400} = \sqrt{2000}$$|\vec{\omega}_R| = 20\sqrt{5}\ rad/s$The resultant vector $\vec{\omega}_R$ lies in a plane making an angle $	heta$ with the horizontal,where $	heta = 	an^{-1}(\frac{\omega_T}{\omega_F}) = 	an^{-1}(\frac{20}{40}) = 	an^{-1}(0.5)$.

$A$ table rotates about a vertical axis passing through its center with an angular velocity of $20\ rad/s$. $A$ flywheel is mounted on it,which rotates about a horizontal axis with an angular velocity of $40\ rad/s$. The resultant angular velocity of the flywheel is:

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