a horizontal disc rotates with a constant angular velocity about a vertical axis passing through

1. The body m experiences a centrifugal force due to the rotation of the disc. This force acts r

A horizontal disc rotates with a constant angular velocity...

A horizontal disc rotates with a constant angular velocity ω about a vertical axis passing through its center. A small body m moves along a diameter with a velocity v. Find the force the disc exerts on the body when it is at a distance r located from the rotation axis. A. mrω^2 + 2mvω B. mg + √(m^2r^2ω^4 + (2mvω)^2) C. √(m^2g^2 + (2mvω)^2) + mrω^2 D. m√(g^2 + r^2ω^4 + (2vω)^2)

The disc $O B A C$ is rotating with angular velocity $\omega$ about the axis $O O^{\prime}$ passing through the edge point $O$. The equation of motion in rotating frame is, $m \vec{w}=\vec{F}+m \omega^{2} \vec{R}+2 m \vec{v} \times \vec{\omega}=\vec{F}+\vec{F}_{i n}$ where $\vec{F}_{i n}$ is the resultant inertial forc ( pseudo force) which is the vector sum of centrifugal and Coriolis forces. (a) At $A, F_{i n}$ vanishes. Thus $0=-2 m \omega^{2} R \hat{n}+2 m v^{\prime} \omega$ where $\hat{n}$ is the inward drawn unit vector to the centre from the point in question, here $A$. Thus, $\quad v^{\prime}=\omega R$ so, $w=\frac{v^{\prime 2}}{\rho}=\frac{v^{\prime 2}}{R}=\omega^{2} R$ (b) At $B$ $\vec{F}_{i n}=m \omega^{2} \overrightarrow{O C}+m \omega^{2} \overrightarrow{B C}$ its magnitude is $m \omega^{2} \sqrt{4 R^{2}-r^{2}}$, where $r=O B$.

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