A thin disk of mass 2.2 kg and radius 61.2 cm is suspended by a horizonal axis perpendicular to the disk through its rim. The disk is displaced slightly from equilibrium and released. The acceleration of gravity is 9.81 m/s2. Find the period of the subsequent simple harmonic motion. Answer in units of s.

The period of the subsequent simple harmonic motion is 1.004 sec.

Explanation:

The given data is as follows.

  Mass of disk (m) = 2.2 kg,     radius of the disk (r) = 61.2 cm,

Formula to calculate the moment of inertia around the center of mass is as follows.

                     =

                     = 0.412

Also,

Distance between center of mass and axis of rotation (d) = r = 0.612 m

Moment of inertia about the axis of rotation (I)

        I =

       I =

          = 0.339

Now, we will calculate the time period as follows.

        T =

        T =

        T = 1.435 sec

      T =

          = 1.004 sec

Thus, we can conclude that the period of the subsequent simple harmonic motion is 1.004 sec.

snell's law

refractve index = sin incidence / sin refraction

k2s for potassium sulphide