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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:

enter image description here

$$\vec{L_P} = m(\vec{r} \times \vec{v_{com}}) + I_{com} \vec{\omega}$$

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?

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  • $\begingroup$ I've removed some comments that were providing (or attempting to provide) answers to the question. Please post a proper answer if you have one and use comments to suggest/request clarifications in the post. Thanks! $\endgroup$ – tpg2114 Oct 6 '20 at 12:52
  • $\begingroup$ curl your fingers in the direction of rotation and see where your thumb points , the error you made is in that part. $\endgroup$ – JustJohan Oct 6 '20 at 12:55
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    $\begingroup$ What is your definition of $\vec{r}$ and is possible you have it flipped around? $\endgroup$ – John Alexiou Oct 6 '20 at 12:55
  • $\begingroup$ Related? physics.stackexchange.com/a/395043/392 $\endgroup$ – John Alexiou Oct 6 '20 at 12:56
  • $\begingroup$ @JohnAlexiou Thank You! That's where I went wrong. A silly question. $\endgroup$ – Arnav Mahajan Oct 6 '20 at 14:24
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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.

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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.

diagram

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 $\omega$.

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