Kinetic rotational energy at different rotational axis

Question

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.

1. 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 $\frac{3}{2}mr^2$

2. 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.

3. 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.

Answer 1

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 .

why is it different along a different axis

there is no different , the kinetic energy is the same !

Answer 2

No, you can very well calculate the kinetic energy of any rigid system about any point, although some derivation is required. KE = (summation) 1/2 (Mi) (Vi)^2 where Mi and Vi are mass of velocity of ith particle. We are summing all such particles for complete KE Let us sit on a point P to calculate the KE. point P moves with velocity V and the body rotates with angular velocity w about it.

For a point i at a distance (Ri) from P, the velocity of ith particle is (V + (Ri)x(w))

(this is in vector notation, x refers to cross product, im new to stack exchange and dont know yet how to use fancy symbols)

KE = (summation)(1/2 (Mi) (V + (Ri)x(w))^2

KE =(summation)(1/2 (Mi) (v^2 + w^2 (Ri)^2 + 2v*(Ri)w)

KE = (summation)(1/2 (Mi) v^2 + 1/2 (Mi)(Ri)^2 w^2 + (Mi)w(v*(Ri))

see that the first term after summation is just the KE of body as if the entire system were condensed onto point P and the second term is 1/2 (Ip) w^2 where Ip is moment of inertia about point P

KE = 1/2 M v^2 + 1/2 (Ip) w^2 + v* (summation)(Mi)(Ri*w))

KE = 1/2 M v^2 + 1/2 (Ip) w^2 + v* w*(summation)(Mi)(Ri)) as w is constant

KE = 1/2 M v^2 + 1/2 (Ip) w^2 + wM(v((summation)(Mi)(Ri))/M), we have divided and mutiplied by M,

notice how (summation)(Mi)(Ri))/M is just position of COM wrt to point P lets denote this with (Rcomp)

hence , in final form:

KE about any point P, whose velocity is V, on the rigid body is

KE = 1/2 M V^2 + 1/2 (Ip) w^2 + mw(V*(Rcom))

V*(Rcom) is a dot product and should be taken with velocity vector V of point P and position vector from point P to COM

Notice how this equation becomes normal KE equation of we take P as COM (third term = 0)

Sorry if the explanation is a little jumbled up and unorganised :/