1.

A metal piece of mass 120g is stretched to form a plane rectangular sheet of area of cross section 0.54m^(2). If length and breadth of this sheet are in the ratio 1:6, find its moment of inertia about an axis passing through its centre and perpendicular to its plane.

Answer»

SOLUTION :Mass `M=120g=120xx10^(-3)kg`
area `=lb=0.54m^(2)`
`(l)/(b)=(1)/(6)` `:.l=(b)/(6)`
`lb=0.54`, `(b)/(6).b=0.54`
`b^(2)=0.54xx6impliesb=sqrt(3.24)=1.8m`
SIMILARLY `l=(0.54)/(1.8)=0.3m`
Moment of Inertia
`I=(M(l^(2)+b^(2)))/(12)=(120xx10^(-3)[(0.3)^(2)+(1.8)^(2)])/(12)`
`I=33.3xx10^(-3)kgm^(2)`


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