TS EAMCET · Physics · Rotational Motion
Assume the earth's orbit around the sun as circular and the distance between their centres as \(D\). Mass of the earth is \(M\) and its radius is \(R\). If earth has an angular velocity \(\omega_0\) with respect to its centre and \(\omega\) with respect to the centre of the sun, the total kinetic energy of earth is :
- A \(\frac{M R^2 \omega_0^2}{5}\left[1+\left(\frac{\omega}{\omega_0}\right)^2+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]\)
- B \(\frac{M R^2 \omega_0^2}{5}\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]\)
- C \(\frac{2}{5} M R^2 \omega_0^2\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]\)
- D \(\frac{2}{5} M R^2 \omega_0^2\left[1+\left(\frac{\omega}{\omega_0}\right)^2+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]\)
Answer & Solution
Correct Answer
(B) \(\frac{M R^2 \omega_0^2}{5}\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]\)
Step-by-step Solution
Detailed explanation
\((\mathrm{KE})_{\text {total }}=\frac{1}{2} \times \frac{2}{5} M R^2 \omega_0^2+\frac{1}{2} M R^2\left(\frac{D^2}{R^2}\right) \omega^2\) \(=\frac{M R^2 \omega_0^2}{5}\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]\)
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