TS EAMCET · Physics · Rotational Motion
A rod of length \(L\) revolves in a horizontal plane about the axis passing through its centre and perpendicular to its length. The angular velocity of the \(\operatorname{rod}\) is \(\omega\). If \(A\) is the area of cross-section of the rod and \(\rho\) is its density, then the rotational kinetic energy of the rod is
- A \(\frac{1}{3} A L^3 \rho \omega^2\)
- B \(\frac{1}{2} A L^3 \rho \omega^2\)
- C \(\frac{1}{24} A L^3 \rho \omega^2\)
- D \(\frac{1}{18} A L^3 \rho \omega^2\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{24} A L^3 \rho \omega^2\)
Step-by-step Solution
Detailed explanation
Rotational kinetic energy of rod is given as \(K_{\mathrm{rot}}=\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{M L^2}{12}\right) \omega^2\) \(K_{\mathrm{rot}}=\frac{1}{24} M L^2 \omega^2\) \(\ldots(\mathrm{i})\) But \(\quad M=\) volume of rod \(\times\) density…
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