Respuesta :
Answer:
0.0651 kgm²
Explanation:
Parameters given:
Mass of meter stick, m = 0.39 kg
Distance of axis from center of meter stick, D = 50 - 21.1 cm = 28.9 cm = 0.289 m
The moment of inertia of a thin rod about an axis that is perpendicular to it and is located at a particular distance, D, from its center is given as:
I = 1/12 * m * L² + m * D²
I = (1/12 * 0.39 * 1²) + (0.39 * 0.289²)
I = 0.325 + 0.326
I = 0.0651 kgm²
The rotational inertia of a meter stick is [tex]0.00423 \;\rm kgm^2[/tex].
Given that, the mass of the meter stick is 0.390 kg.
The moment of inertia of the rod about its central axis is calculated by the formula given below.
[tex]I = \dfrac {M l^2} {3}[/tex]
Where [tex]I[/tex] is the moment of inertia of the rod, [tex]M[/tex] is the mass of the rod and [tex]l[/tex] is the length of the rod.
If the distance of the perpendicular axis from the center of the stick is 21.1 cm. Then as per the parallel axis theorem, the moment of inertial of the stick can be calculated by the formula given below.
[tex]I_1 = I + M d^2[/tex]
Where [tex]I_1[/tex] is the moment of inertial of the stick about the perpendicular axis, [tex]I[/tex] is the moment of inertia of the stick and [tex]d[/tex] is the distance of the perpendicular axis form the center of the stick.
So [tex]I_1 = \dfrac {1}{3} \times M l^2 + M d^2[/tex]
As given that M = 0.390 Kg, l = 1 meter and d = (0.50 - 0.211) m = 0.289 m.
[tex]I_1 = \dfrac {1}{3} \times (0.390\times 1^2) + 0.390 \times (0.289)^2[/tex]
By simplifying the above equation, we get
[tex]I_1 = 0.00423[/tex]
The rotational inertia of a meter stick is [tex]0.00423 \;\rm kgm^2[/tex].
For more details, follow the link given below.
https://brainly.com/question/6953943.
