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Four spheres each of diameter 0.02m and 0.10 kg are placed with their centres on the vertices of a square of side 0.05m. Calculate the moment of inertia of the system about one side of the taken as the axis of rotation. |
Answer» Solution : Let the AXIS of rotation be ALONG AD. MOMENT of inertia of B and C about the axis XY `=I_0+Md^2` i.e M.I`=2/5MR^2+Md^2=0.1[0.4xx(0.01)^2+(0.05)^2]` M.I`=2.54xx10^(4)kgm^2` Moment of inertia of A & D about an axis XY is `=2/5MR^2` `=0.4xx0.1xx(0.01)^2` `=4xx10^(-6)=0.04xx10^(-4)kgm^2` Hence total M.l of the system`=2xx2.54xx10^(-4)+2xx0.04xx10^(-4)` `=5.16xx10^(-4)kgm^2` |
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