Undergrad Energy of spinning objects as axis of rotation moves

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SUMMARY

The discussion centers on the dynamics of spinning objects, specifically how the axis of rotation converges to the center of mass (CM) to minimize the moment of inertia (I). The equations presented, including $$E=\frac{1}{2} I \omega^2$$ and $$\frac{dE}{dt}=\frac{dI}{dt}\omega^2+ 2I\omega\frac{d\omega}{dt}=0$$, illustrate that as I decreases, angular velocity ($$\omega$$) increases, leading to a change in rotational energy. The total kinetic energy of a rigid body is expressed as $$E_{\text{kin}}=T=\frac{M}{2} \dot{\vec{R}}^2 + \frac{1}{2} \vec{\omega} \hat{\Theta} \vec{\omega}$$, emphasizing the importance of both translational and rotational components in energy calculations.

PREREQUISITES
  • Understanding of rotational dynamics and angular momentum
  • Familiarity with the concepts of moment of inertia and center of mass
  • Basic knowledge of differential equations and their applications in physics
  • Proficiency in kinetic energy calculations for rigid bodies
NEXT STEPS
  • Study the principles of angular momentum conservation in non-rigid bodies
  • Explore the derivation and applications of the inertia tensor in three-dimensional motion
  • Learn about the effects of external forces on the rotational motion of objects
  • Investigate the relationship between translational and rotational kinetic energy in complex systems
USEFUL FOR

Physicists, mechanical engineers, and students studying dynamics who are interested in the behavior of spinning objects and the interplay between rotational and translational energy.

Trollfaz
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Imagine an object, e.g throwing knife, spins in the air but not forced to rotate about a particular axis, i.e no rod impaling it and forcing it to spin about the rod. Then the axis of rotation converges to it's center of mass (CM) to minimize I. But there's nowhere for it's rotation energy to go assuming no resistive forces of the medium.
$$E=\frac{1}{2} I \omega^2= k$$
Both I and ##\omega## are functions of t and
##\frac{dI}{dt}<0##
So angular velocity increases?
$$\frac{dE}{dt}=\frac{dI}{dt}\omega^2+ 2I\omega\frac{d\omega}{dt}=0$$
$$\frac{d\omega}{dt}=-\frac{dI}{dt}\omega^2/2I\omega>0$$
And if we know the rate of change of I we can solve this differential equation to find how angular velocity evolves
 
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Your expression for the energy only describes the rotational energy of the body around the given axis but not the translational energy of the body as a whole. It's most simple to choose the center of mass as the body-fixed reference point. Then the total kinetic energy of the rigid body reads
$$E_{\text{kin}}=T=\frac{M}{2} \dot{\vec{R}}^2 + \frac{1}{2} \vec{\omega} \hat{\Theta} \vec{\omega},$$
where ##M## is the total mass and ##\hat{\Theta}## is the tensor of inertia around the center of mass of the body.
 

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