Answer:
[tex]I=0.0503\ kg-m^2[/tex]
Explanation:
Given that,
Mass of the object, m = 29 kg
Torsion constant of the wire, K = 1.14 N-m
Number of cycles, n = 98
Time, t = 74 s
To find,
The rotational inertia of the object.
Solution,
Relationship between the moment of inertia, time period and the torsion constant of the spring is given by :
[tex]T=2\pi\sqrt{\dfrac{I}{K}}[/tex]
Where I is the moment of inertia
K is spring constant
Let T Is the time period of oscillation, such that,
[tex]T=\dfrac{98}{74}=1.32\ s[/tex]
[tex]I=\dfrac{T^2K}{4\pi ^2}[/tex]
[tex]I=\dfrac{(1.32)^2\times 1.14}{4\pi ^2}[/tex]
[tex]I=0.0503\ kg-m^2[/tex]
So, the rotational inertia of the object is [tex]0.0503\ kg-m^2[/tex].
