Respuesta :
Answer:[tex]I=3.024\ kg-m^2[/tex]
Explanation:
Given
mass of first [tex]m_1=5 kg[/tex]
mass of second [tex]m_2=3 kg[/tex]
Distance of [tex]m_1[/tex] from origin [tex]r_1=36 cm[/tex]
Distance of [tex]m_2[/tex] from origin [tex]r_2=89 cm[/tex]
Moment of inertia is given by multiplication of mass and square of distance
[tex]I=\sum mr^2[/tex]
[tex]I=m_1r_1^2+m_2r_2^2[/tex]
[tex]I=5\times 0.36^2+3\times 0.89^2[/tex]
[tex]I=0.648+2.376[/tex]
[tex]I=3.024\ kg-m^2[/tex]
Answer:
[tex]I=4.47\ kg.m^{2}[/tex]
Explanation:
Given:
A light meter stick (assuming mass-less) is loaded with point masses at the following positions.
- mass of first object, [tex]m_1=5\ kg[/tex]
- mass of second object, [tex]m_2=3\ kg[/tex]
- position of the first mass, [tex]x_1=36\ cm=0.36\ m[/tex]
- position of the second mass, [tex]x_2=89\ cm=0.89\ m[/tex]
As we know that moment of inertia for point mass is given by:
[tex]I=m.R^2[/tex]
where:
[tex]m=[/tex] mass of the object
[tex]R=[/tex] radial distance from the axis of rotation
Now the total moment of inertia:
[tex]I=m_1.x_1^2+m_2.x_2^2[/tex]
[tex]I=5\times 0.36+3\times 0.89[/tex]
[tex]I=4.47\ kg.m^{2}[/tex]
