Kinetic Energy of a rigid body and Instant centre of rotation [duplicate]

Question

In my physics textbook the kinetic energy of a rigid body that rotates around a point( like a pendulum) is derived this way:

$$\sum_{n=1}^{N}\dfrac{1}{2}m_nv^2_n = \sum_{n=1}^{N}\dfrac{1}{2}m_n(\omega r)^2=\sum_{n=1}^{N}\dfrac{1}{2}m_n(\omega r_n)^2=\dfrac{1}{2}I\omega ^2 $$

where $\omega$ is angular velocity, and r is the radius.

This last member of this equation doesn't have trace of linear velocity, I suppose because from the chosen frame of reference the object is only rotating, by applying Huygens-Steiner we get: $ \dfrac{1}{2}I\omega ^2 $=$\dfrac{1}{2}(I_C+Md^2)\omega ^2 $=$\dfrac{1}{2}I_C\omega ^2+\dfrac{1}{2}(Md^2)\omega ^2$=$\dfrac{1}{2}I_C\omega ^2+\dfrac{1}{2}MV_C ^2$

where Vc is the velocity of center of mass and Ic is the momentum of inertia and d the distance of the center of mass.

In this case we do have a 'linear' velocity, since from this frame of reference we have the center of mass that isn't rotating.

My question is: Can we always calculate the energy of a rigid body with this $\dfrac{1}{2}I\omega ^2 $ , even when the body isn't rotating around a fixed point( an example could be a ball that got kicked and is moving and rotating)? My idea was to use an axis that goes through the Instant centre of rotation so we would only see a pure rotation. Is it more convenient to use this version of the formula $\dfrac{1}{2}I_C\omega ^2+\dfrac{1}{2}MV_C ^2$ ?

Answer 1

with configuration I

$$v_1=\omega\,r_1\\ v_2=\omega\,r_2\\ \vdots\\ v_n=\omega\,r_n$$

the kinetic energy $$T:=\frac 12\sum m_n\,v_n^2= \frac 12\,\omega^2\underbrace{\sum m_n\,\,r_n^2}_{I}=\frac 12\,I\,\omega^2\quad n=1\ldots\infty$$

with the center of mass coordinates configuration II

$$I=I_C+M\,d^2\quad,v_c=\omega\,d\quad\Rightarrow$$ $$\frac 12\,I\,\omega^2=(I_C+M\,d^2)\,\omega^2= \frac 12\left[I_C\,\omega^2+M\,d^2\underbrace{\left(\frac {v_c}{d}\right)^2}_{\omega^2}\right]=\frac 12\,I_C\,\omega^2+\frac 12\,M\,v_c^2$$

where $~M=\sum m_n~$

Can we always calculate the energy of a rigid body with this ...

this equation

$$T=\frac 12\,I_C\,\omega^2+\frac 12\,M\,v_c^2$$

is only valid if the angular velocity vector $~\vec\omega~$ is scalar

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the kinetic energy for a rigid body is

$$T=\frac 12\,M\,\vec v_c\cdot \vec v_c+\frac 12 \vec\omega^T\,\mathbf I_{C}\,\vec\omega$$

where $~I_C~$ is the inertia tensor and the angular velocity $~\vec \omega~$ is a function of the body rotation angles and rotation angles velocities .