A disk of radius R, mass M and uniform density and thickness is pivoted at point

Question

A disk of radius R, mass M and uniform density and thickness is pivoted at point A at the disk's edge. The disk swings left ~and~right on the pivot about the equilibrium position directly below the pivot point A. Calculate the period of small oscillations of the disk. Use small angle approximation.

1. T = 2pi sqrt(5R/3g)

2. T = 2pi sqrt(R/3g)

3. T = 2pi sqrt(R/2g)

4. T = 2pi sqrt(3R/2g)

5. T = 2pi sqrt(R/g)

6. T = 2pi sqrt(4R/3g)

7. T = 2pi sqrt(2R/3g)

8. T = 2pi sqrt(11R/6g)

9. T = 2pi sqrt(2R/g)

10. T = 2pi sqrt(7R/6g)

Explanation / Answer

The period of a physical pendulum for small angle of oscillations is
(1) ..... T = 2*?*sqrt [ I /(M*g* D ) ]

where R is the distance of the mass M from the axis of rotation and I is
the moment of inertia of the disk about the axis of rotation. Applying
the parallel axis theorem to a disk with radius R whose axis
of rotation passes on the rim perpendicular to the plane of the disk

(2) ..... I = Icm + M*R ^2  = (1/2)*M*(R^2) + M*R ^2

.......... I = (1/2)*M*[ (R^2) + 2*(R ^2) ] = (1/2)*M*[ 3*(R ^2) ]

.......... I/(M*g* R ] = (3/2)*M*(R^2) /(M*g* R ] = (3/2)*R /g = 3*R /(2*g)

so that (1) becomes

(3) ..... T = 2*?*sqrt [3*R /(2*g)] (Answer)

So point 4 is correct.

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