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.
[tex]I_{cm} = \frac{1}{2}mr^{2}[/tex]
= [tex]\frac{1}{2} \times 2.2 kg \times (61.2)^{2}[/tex]
= 0.412 [tex]kg m^{2}[/tex]
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 = [tex]I_{cm} + md^{2}[/tex]
I = [tex]0.412 + (2.2) \times (0.612)^{2}[/tex]
= 0.339 [tex]kgm^{2}[/tex]
Now, we will calculate the time period as follows.
T = [tex]2\pi sqrt{\frac{I}{mgd})[/tex]
T = [tex]2 \times 3.14 sqrt{(\frac{0.339}{2.2 \times 9.8 \times 0.612}[/tex]
T = 1.435 sec
T = [tex]2 \times 3.14 \times 0.16[/tex]
= 1.004 sec
Thus, we can conclude that the period of the subsequent simple harmonic motion is 1.004 sec.
