KCET · Physics · Rotational Motion
A thin uniform rectangular plate of mass \(2 \mathrm{~kg}\) is placed in \(x y\)-plane as shown in figure. The moment of inertial about \(x\)-axis is \(I_{x}=0.2 \mathrm{kgm}^{2}\) and the moment of inertia about \(Y\)-axis is \(I_{y}=0.3 \mathrm{kgm}^{2}\). The radius of gyration of the plate about the axis passing through \(O\) and perpendicular to the plane of the plate is
- A \(50 \mathrm{~cm}\)
- B \(5 \mathrm{~cm}\)
- C \(38.7 \mathrm{~cm}\)
- D \(31.6 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(A) \(50 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Given, mass, \(M=2 \mathrm{~kg}\)
\(\begin{aligned}
&I_{x}=0.2 \mathrm{~kg}-\mathrm{m}^{2} \\
&I_{y}=0.3 \mathrm{~kg}-\mathrm{m}^{2}
\end{aligned}\)
\(\text { and } \quad I_{y}=0.3 \mathrm{~kg}-\mathrm{m}^{2}\)
According to the perpendicular axis theorem, the moment of inertia of the rectangular plate about an axis passing through \(O\) and perpendicular to the plane of plate is
\(I=I_{x}+I_{y}=0.2+0.3\)
\(=0.5 \mathrm{~kg} \mathrm{~m}{ }^{2}\)
Also, \(I=M k^{2}\)
where, \(k=\) radius of gyration.
\(\Rightarrow k=\sqrt{\frac{I}{M}}=\sqrt{\frac{0.5}{2}}=\sqrt{0.25}=0.5 \mathrm{~m}\) or \(50 \mathrm{~cm}\)
\(\begin{aligned}
&I_{x}=0.2 \mathrm{~kg}-\mathrm{m}^{2} \\
&I_{y}=0.3 \mathrm{~kg}-\mathrm{m}^{2}
\end{aligned}\)
\(\text { and } \quad I_{y}=0.3 \mathrm{~kg}-\mathrm{m}^{2}\)
According to the perpendicular axis theorem, the moment of inertia of the rectangular plate about an axis passing through \(O\) and perpendicular to the plane of plate is
\(I=I_{x}+I_{y}=0.2+0.3\)
\(=0.5 \mathrm{~kg} \mathrm{~m}{ }^{2}\)
Also, \(I=M k^{2}\)
where, \(k=\) radius of gyration.
\(\Rightarrow k=\sqrt{\frac{I}{M}}=\sqrt{\frac{0.5}{2}}=\sqrt{0.25}=0.5 \mathrm{~m}\) or \(50 \mathrm{~cm}\)
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