A uniform disk of mass M and radius R is oriented in a vertical plane. The y axis is vertical, and the x axis is horizontal. The disk is pivoted about the origin of the coordinate system, with the center of the disk hanging a distance \frac{R}{\sqrt{2}} below the pivot, as shown. The disk is free to rotate without friction about the pivot in the x-y plane.

Engineering

Physics

Moment of Inertia

Question

A uniform disk of mass M and radius R is oriented in a vertical plane. The y axis is vertical, and the x axis is horizontal. The disk is pivoted about the origin of the coordinate system, with the center of the disk hanging a distance $\frac{R}{\sqrt{2}}$ below the pivot, as shown. The disk is free to rotate without friction about the pivot in the x-y plane.

the shown position is a position of stable equilibrium.

if disc is rotated by angle 45° and released, its initial angular acceleration is $\frac{g}{2 R}$ .

In the position shown, pivot exerts no force on the disc.

the moment of inertia of disc about pivot is MR²

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I = $\frac{{mR}^{2}}{2}$ + mh² = mR²
𝛕 = h mg sin 45° = Iα
$\frac{R}{\sqrt{2}}$ mg $\frac{1}{\sqrt{2}}$ = mR²α
α = $\frac{g}{2 R}$ ⟹ F = mg

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