Rotation Ques 26
- Four solid spheres each of diameter $\sqrt{5} cm$ and mass $0.5 kg$ are placed with their centres at the corners of a square of side $4 cm$. The moment of inertia of the system about the diagonal of the square is $N \times 10^{-4} kg-m^{2}$, then $N$ is
(2011)
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Answer:
Correct Answer: 26.(9)
Solution:
Formula:
Two Important Theorems On Moment Of Inertia :
- $r=\frac{d}{2}=\frac{\sqrt{5}}{2} cm=\frac{\sqrt{5}}{2} \times 10^{-2} m \Rightarrow m=0.5 kg$
$I _{X X}=I _1+I _2+I _3+I _4$
$=\frac{2}{5} m r^{2}+m \frac{a}{\sqrt{2}}^{2}+\frac{2}{5} m r^{2}$
$+\frac{2}{5} m r^{2}+m \frac{a}{\sqrt{2}}^{2}+\frac{2}{5} m r^{2}$
Substituting the values, we get
$$ I _{X X}=9 \times 10^{-4} kg-m^{2} \therefore \quad N=9 $$
BETA
