ExamBro
ExamBro
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 :

  1. A \(\frac{M R^2 \omega_0^2}{5}\left[1+\left(\frac{\omega}{\omega_0}\right)^2+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]\)
  2. B \(\frac{M R^2 \omega_0^2}{5}\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]\)
  3. C \(\frac{2}{5} M R^2 \omega_0^2\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]\)
  4. D \(\frac{2}{5} M R^2 \omega_0^2\left[1+\left(\frac{\omega}{\omega_0}\right)^2+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]\)
Verified Solution

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

\((\mathrm{KE})_{\text {total }}=\frac{1}{2} \times \frac{2}{5} M R^2 \omega_0^2+\frac{1}{2} M R^2\left(\frac{D^2}{R^2}\right) \omega^2\) \(=\frac{M R^2 \omega_0^2}{5}\left[1+\frac{5}{2}\left(\frac{D \omega}{R \omega_0}\right)^2\right]\)
Same subject
Explore more questions on app
From TS EAMCET
Explore more questions on app