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A system consists of two identical spheres each of  mass 1.5 \mathrm{~kg}$ and radius $50 \mathrm{~cm} at the ends of a light rod. The distance between the centres of the two spheres is 5 \mathrm{~m}. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through it midpoint?
 
Option: 1 1.905 \times 10^{5} \mathrm{kgm}^{2}
Option: 2 18.75 \mathrm{kgm}^{2}
Option: 3 19.05 \mathrm{kgm}^{2}
Option: 4 1.875 \times 10^{5} \mathrm{kgm}^{2}

Answers (1)

best_answer

R=25cm=\frac{1}{4}m 

I_{2}=I_{1}=\left ( \frac{2}{5}MR^{2} \right )= \frac{2}{5}M\times \left ( \frac{1}{4} \right )^{2}\\

I_{0}=\left ( I_{1}+Md^{2} \right )+\left ( I_{2}+Md^{2} \right )\\

     =\frac{4}{5}M\left ( \frac{1}{4} \right )^{2}+2M\times\left ( \frac{10}{4} \right )^{2}\\                  

\because d=2.5m\\

     =\frac{4}{5}\times1.5\times\frac{1}{16}+2\times1.5\times\frac{100}{16}\\

I_{0}=\frac{6}{80}+\frac{1500}{16\times5}=\frac{1506}{80}=18.75kgm^{2}

 

 

Posted by

vishal kumar

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