JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A thin uniform rod of length \(2\,m\). cross sectional area ' \(A\) ' and density ' \(d\) ' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity \(\omega\). If value of \(\omega\) in terms of its rotational kinetic energy \(E\) is \(\sqrt{\frac{\alpha E}{ Ad }}\) then the value of \(\alpha\) is \(...........\)
- A \(2\)
- B \(1\)
- C \(4\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(( KE )_{\text {Rotational }}=\frac{1}{2} I \omega^2= E\) \(E =\frac{1}{2} \frac{ m \ell^2}{12} \omega^2\) \(E =\frac{1}{2} \frac{ dA \ell^3}{12} \omega^2\) \(E =\frac{ dA (2)^3}{24} \omega^2\) \(\sqrt{\frac{3 E }{ dA }}=\omega\) \(\alpha=3 \text { Ans. }\)
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