Question

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.

Answer 1

Answer: 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.

          $I_{cm} = \frac{1}{2}mr^{2}$  

                     = $\frac{1}{2} \times 2.2 kg \times (61.2)^{2}$

                     = 0.412 $kg m^{2}$

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 = $I_{cm} + md^{2}$

       I = $0.412 + (2.2) \times (0.612)^{2}$

          = 0.339 $kgm^{2}$

Now, we will calculate the time period as follows.

        T = $2\pi sqrt{\frac{I}{mgd})$

        T = $2 \times 3.14 sqrt{(\frac{0.339}{2.2 \times 9.8 \times 0.612}$

        T = 1.435 sec

      T = $2 \times 3.14 \times 0.16$

          = 1.004 sec

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