Question 36 of 280

Two identical solid spheres each of mass $2kg$ and radii $10cm$are fixed at the ends of a light rod. The separation between the centres of the spheres is $40cm.$ The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is______\times {10}^{–3}kg{m}^{2}.

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Answer: 176

The moment of inertia of a sphere is given by the formula $I=\dfrac{2}{5}M{R}^{2}$.
Since there are two masses, the total moment of inertia about the axis perpendicular to the rod passing through the centre is
$I=2\times \left[\dfrac{2}{5}M{R}^{2}+M{\left(\dfrac{l}{2}+r\right)}^{2}\right]...(i)$
The given data is
$M=2kg l=\left(0.4-0.2\right)m=0.2m R=0.1m$
Substituting the values in equation (i)
$I=\left(\dfrac{4}{5}(2\times 0.{1}^{2})+2\times 2(\dfrac{0.2}{2}+0.1{)}^{2}\right)kg{m}^{2} \Rightarrow I=\left(\left(\dfrac{4}{5}\times 0.02\right)+\left(4\times 0.04\right)\right)kg{m}^{2} \Rightarrow I=\left(0.016+0.16\right)kg{m}^{2}=0.176kg{m}^{2}$