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How to calculate kinetic energy of rod which is rotating about an external axis and spinning about its centre of mass?

1) I know how to calculate kinetic energy of a rod if it is not spinning around centre of mass but rotating around external axis

length=l

mass=m

angular speed $\omega=\sqrt{\frac{g}{l}}$

non spinnning rod

$$KE=KE_{translation}+KE_{rotation}$$

$$KE=\frac{1}{2}mv_{cm}^2+\frac{1}{2}I_{cm}\omega^2$$

$$v_{cm}=\frac{3l\omega}{2}, I_{cm}=\frac{ml^2}{12}$$

$$KE=\frac{7}{6}mgl$$

$$or$$

$$KE=\frac{1}{2}I\omega^2$$

I is about axis of rotation

$$I=\frac73 ml^2$$

$$KE=\frac{1}{2}(\frac73 ml^2)\frac{g}{l}$$

$$KE=\frac{7}{6}mgl$$

But now if rod is spinning around its com as well as external axis

2) A rod is spinning as shown

in this case im getting trouble

enter image description here

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  • $\begingroup$ just add the two energies $\endgroup$ – trula Apr 25 at 14:20
  • $\begingroup$ In the first case, rod is not rotating about its own COM, then it's KE should be zero about COM, but we taken KE about COM by considering angular speed w(omega). $\endgroup$ – teja Apr 25 at 14:53
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    $\begingroup$ If the first rod is attached to the axle by a cord, then the angular velocity about the COM is the same as about the axle. $\endgroup$ – R.W. Bird May 1 at 20:01
  • $\begingroup$ R.W. Bird , can you explain why angular velocities are equal if cord is attached $\endgroup$ – teja May 3 at 1:17
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    $\begingroup$ By combined motion, (refer physics.stackexchange.com/questions/546718/…), the answer is 1/2M(3l/2∗w2)²+1/2ML2/12(w21+w22), as w net has components w1 and w2, along two mutually perpendicular axis $\endgroup$ – user600016 May 17 at 2:35
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To calculate the kinetic energy of the rod with respect to the external spinning axis, consider two parts of the kinetic energy:

1) rotational energy of the entire rod with respect to the external axis (view the entire rod as a point of mass at its center of mass): $$KE_1 = \frac 1 2 (m (\frac {3l} 2)^2)\omega_2^2$$ 2) rotational energy of the rod with respect to its center of mass: $$KE_2 = \frac 1 2 I \omega_1^2$$ where $I$ is the moment of inertia of the rod with respect to its center of mass.
Adding them together gives the total kinetic energy.

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  • $\begingroup$ This answer is incorrect. By combined motion, (refer physics.stackexchange.com/questions/546718/…), the answer is $1/2M(3l/2*w_2)² + 1/2 ML^2/12 ( w_1^2+w_2^2)$, as w net has components w1 and w2, along two mutually perpendicular axis. $\endgroup$ – user600016 May 17 at 2:35
  • $\begingroup$ @user600016 But I failed to see any difference between the expression you gave and that is in my answer... $\endgroup$ – user12986714 May 17 at 3:31
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    $\begingroup$ KE2 should be $\frac{1}{2} I (w_1^2+w_2^2)$. Sorry for not formating the expression in the previous comment properly $\endgroup$ – user600016 May 17 at 4:23

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