IMP PTS REGARDING MOMENT OF INERTIA
- MOMENT OF INERTIA is rotation analogue of mass
- Property due to which a body resists any change in rotational stat
- MOI depends on mass as well as its distribution wrt axis of rotation
- MOI I=mr^2 where r=perpendicular distance of pt mass m frm axis of rotation. So if the axis inclined at
frm vertical I=mr^2sin^2
- MOI I= dmr^2 for rigid body
- I=0 for pt mass on the axis of rotation
- theorem of parll axis(for both 3D and 2D):-MOI abt an axis passing though COM of body at distance r is given by I=Icm+mr^2
- theorem of perpedicular axis(app only for planar bodies):-for a body in XY plane Iz=Ix+Iy
- MOI can be added or subtracted when written WRT SAME AXIS
- MOI abt symmetric axis will be same.Eg MOI wrt the two diagnols of square will be equal
- Projection of a body's inertia and formed shape are equal .Eg: MOI of thin rod and square sheet(elongated rod) will be same abt the same axis
- If a part is removed Iremaining=Iwhole body-Iremoved part abt same axis
- MOI of segment=(mass of segment)*r^2
- DIRECT BUT IMP RESULTS
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- MOI abt any line in the plane of a square lamina passing through its centre will be same and =Iaxis/2
- For a triangular lamina MOI abt a perpendicular axis passing through its vertices
- Ia=m/12(3b^2+3c^2-a^2) Ib=m/12(3c^2+3a^2-b^2) Ic=m/12*(3a^2+3b^2-c^2)
- MOI for rod Iaxis= 1/12*ml^2
- MOI for thin rectangular sheet abt any diagnol I=ml^2b^2/(l^2+b^2)
- MOI for ring Iperpendicular=mr^2
- MOI for disc Iperpendicular=1/2*mr^2
- MOI for solid sphere I=2/5*mr^2 (abt any axis through centre)
- MOI for hollow sphere I=2/3*mr^2 (abt any axis through centre)
- MOI for solid cylinder(elongated disc) Iaxis=1/2*mr^2 Iequitorial axis=m(l^2/12+r^2/4)
- MOI for hollow cylinder(elongated ring) Iaxis=mr^2 Iequitorial axis=m(l^2/12+r^2/2)
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