Where am I going wrong in finding direction of angular momentum?

Question

If I consider the situation of the problem below, and try to calculate the angular momentum of the rotating (without slipping) solid sphere about point $P$, then obviously I'll use the formula:

$$vec{L_P} = m(vec{r} times vec{v_{com}}) + I_{com} vec{omega}$$
The direction of $vec{omega}$ and $vec{r} times vec{v}$ must be same. As it is a case of pure rolling. But if I try to find the direction of the latter quantity using the right hand palm rule I get it as $+hat{k}$ and if I find that of the former using corkscrew rule I get $-hat{k}$. Why aren't these two directions same? Where do you think I might be going wrong?

Yejus · Accepted Answer

There is no contradiction here. Your direction of the 'spin' angular momentum is incorrect: curling the fingers of your right hand clockwise, you can deduce the direction of $hat{omega}$ is toward $-hat{z}$.
For the 'orbital' angular momentum, the direction is toward $frac{vec{r} times vec{v_{text{com}}}}{|vec{r} times vec{v_{text{com}}|}} = hat{y} times hat{x} = -hat{z},$ which is the same direction as the previous. This means the two angular momenta add constructively.

John Alexiou · Answer

First look at the kinematics. In the case, the contact point has zero velocity, and hence the rotational velocity vector $vec{omega}$ goes into the plane.

Consider the velocity of the center of mass $vec{v}_c = pmatrix{dot{x}_c & 0}$ as well as its location w.r.t. coordinate system origin $vec{r}_c$. The no-slip condition is
$$ hat{i} cdot left( vec{v}_c + (omega hat{k}) times (- R hat{j}) right) = 0 $$
$$ dot{x}_c + R omega = 0 $$
$$ omega = - frac{dot{x}_c}{R} $$
So now lets us look at momentum. Linear momentum is $vec{p}=m pmatrix{dot{x}_c & 0 }$ and angular momentum (in scalar form) about the center of mass is $$ L_c ={rm  I}_c omega = -{rm I}_c frac{dot{x}_c}{R}$$
or in vector form $vec{L}_c = L_c hat{k} = (-{rm I}_c frac{dot{x}_c}{R}) hat{k}$.
Now let us transform this to point P
$$ vec{L}_p = vec{L}_c + vec{p} times ( vec{r}_p - vec{r}_c) $$
$$ vec{L}_p  = (-{rm I}_c frac{dot{x}_c}{R}) hat{k} + (-m R dot{x}_c) hat{k} $$
or in scalar form $$L_p = -left( frac{I_c + m R^2}{R} right) dot{x}_c $$
which is also pointing into the plane, just like $L_c$ and $omega$.

Where am I going wrong in finding direction of angular momentum?

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