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:

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

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?

  • $\begingroup$ 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

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.


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 $\omega$.


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