1.

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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