Consider a homogeneous wheel with a moment of inertia of $\frac{1}{2}mr^2$ around its center of mass. Said wheel rolls along a horizontal area. I am wondering about some of the quantities of this wheel and hope someone can clear things up for me a bit.
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I know that the important rotational axis that causes the motion is not the middle of the wheel, but the connection line of the wheel with the area. So according to the axis theorem I can calculate the moment of inertia about the current rotational axis as $\frac{3}{2}mr^2$
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When considering the rotational kinetic energy with respect to the axis, we get that $E_{kin} = \frac{3}{4}mr^2\omega^2$. This corresponds to the total kinetic rotational energy of the wheel.
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When considering the kinetic energy with respect to the middle of the wheel, we get that the kinetic energy is actually made out of a rotational part ($\frac{1}{4}mr^2\omega^2$) and a translational part $(\frac{1}{2}mr^2\omega^2)$ which sums back up to the total energy of 2.)
My question is, why is it different along a different axis? I understand that the center of mass sees a translation along the horizontal area. Does that mean that energy is defined around a point/axis? Why do we consider a point on the outer egde of the wheel to not be in translation? According to the axis theorem, this would mean that the closer we go inwards, the larger the translational part of the kinetic energy grows, right? I feel like I am missing a very (probably simple) important point of view here.
Best Answer
you have this situation
I)
the kinetic energy at the center of mass is:
$$T=\frac m2\,v_{\rm CM}^2+\frac 12\,I_{\rm CM}\,\omega^2\tag 1$$
If the wheel is rolling with out sliding , this means that
the relative velocity at the contact point A is zero
$$\omega\,r-v_{\rm CM}=0\tag 2$$
with Eq. (2) you obtain
$$T=\frac m2\,(\omega\,r)^2+\frac 12\,I_{\rm CM}\omega^2= \frac 12(m\,r^2+I_{\rm CM})\omega^2\tag 3$$
II)
obtain the momentum about point A (A is the instantaneous rolling point)
$$L=I_A\omega=(m\,r^2+I_{\rm CM})\omega$$
from here the kinetic energy
$$T= \frac 12(m\,r^2+I_{\rm CM})\omega^2$$
the same result as above .
there is no different , the kinetic energy is the same !