Answer:
The rotational inertia of the pendulum around its pivot point is [tex]0.280\,kg\cdot m^{2}[/tex].
Explanation:
The angular frequency of a physical pendulum is measured by the following expression:
[tex]\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }[/tex]
Where:
[tex]\omega[/tex] - Angular frequency, measured in radians per second.
[tex]m[/tex] - Mass of the physical pendulum, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]d[/tex] - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.
[tex]I_{O}[/tex] - Moment of inertia with respect to pivot point, measured in [tex]kg\cdot m^{2}[/tex].
In addition, frequency and angular frequency are both related by the following formula:
[tex]\omega =2\pi\cdot f[/tex]
Where:
[tex]f[/tex] - Frequency, measured in hertz.
If [tex]f = 0.658\,hz[/tex], then angular frequency of the physical pendulum is:
[tex]\omega = 2\pi \cdot (0.658\,hz)[/tex]
[tex]\omega = 4.134\,\frac{rad}{s}[/tex]
From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:
[tex]\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}[/tex]
[tex]I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}[/tex]
Given that [tex]m = 1.15\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]d = 0.425\,m[/tex] and [tex]\omega = 4.134\,\frac{rad}{s}[/tex], the moment of inertia associated with the physical pendulum is:
[tex]I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}[/tex]
[tex]I_{o} = 0.280\,kg\cdot m^{2}[/tex]
The rotational inertia of the pendulum around its pivot point is [tex]0.280\,kg\cdot m^{2}[/tex].
