We know that angular momentum of a solid disk rotating with angular velocity $\omega$ is given by $I\omega$ about its center. Now if I chose any axis (parallel to above), will it have same magnitude of angular momentum as in the above case? If not how should I tackle this axis of rotation?

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    $\begingroup$ Use the parallel axis theorem $\endgroup$ – John Rennie Apr 19 '16 at 6:20
  • $\begingroup$ When referring to moment of inertia abtained by above theorem what will I take angular velocity? I.e its angular velocity about what axis? $\endgroup$ – Muzamil Sheikh Apr 19 '16 at 6:25

When youare talking about angular momentum about any axis(inluding one that passes through center of mass), you carry out the "formula":


where direction of $\vec{L_{com}}$ and $I_{com}\vec{\omega}$ is to be kept in mind, during vector summation.

While we calculate $\vec{L}$, about COM, $\vec{L_{com}}$ becomes zero because, PERPENDICULAR distance between COM and point of reference becomes zero.

Also note that $\vec{L}$ about any axis is NOT equal to $I_{axis}\vec{\omega}$.

This is ONLY true if and only if the body is actually rotating about that axis or the point of contact is instantaneously at rest at that point( i.e. the body is instantaneously rotating about that point, as is the case during pure rolling motion of a ball or body on ground).

Now by parallel axis theomem we can calculate $$I_{axis} = I_{\mathrm{cm}} + md^2$$.

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    $\begingroup$ I don't understand why you are saying the other answers are wrong. It seems like all you have added is that angular momentum is made up of the CM momentum plus the angular momentum around the CM, which isn't really pertinent to this question. $\endgroup$ – wgrenard Apr 19 '16 at 7:16
  • $\begingroup$ Well, whats wrong is wrong. And the addition to my answer is "the answer" to the question. $\endgroup$ – Prayas Agrawal Apr 19 '16 at 7:18
  • $\begingroup$ Yes! this is what I asked. But plz make me more clear that do axis of rotation can be assumed as both moving with object and stationary ? I.e we take axis of rotation in some problems (collisions) moving with object while in some fixed wrt object. $\endgroup$ – Muzamil Sheikh Apr 19 '16 at 7:21
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    $\begingroup$ This question is about a special case, so the other answers are not wrong and shouldn't be downvoted for addressing the special case. If only the most general answers are correct then you should be mentioning principal axes and the inertia tensor, etc. Anyway, that is just my two cents. Good day! $\endgroup$ – wgrenard Apr 19 '16 at 7:23
  • $\begingroup$ Yes you definetly can. But make clear that axis is not acclerating. If it accelerates, then you have to apply psuedo force resulting from accelration of axis, to center of mass $\endgroup$ – Prayas Agrawal Apr 19 '16 at 7:25

When you talk about a rigid body the distance between any two points is fixed. That implies angular velocity w is same for all points. Now ony moment of inertia changes:

I=Icm+md^2 [d is distance between cm and the point]

This is the parallel axis theorem.

Get I and angular momentum=Iw

  • $\begingroup$ Sorry to say but I'm little unclear, suppose in space u are standing and a disk is rotating and moving towards u I choose u as axis of rotation then by by above theorem "d" decreases as disk come closer to u, so angular momentum must decrease withou any agent? $\endgroup$ – Muzamil Sheikh Apr 19 '16 at 7:06
  • $\begingroup$ @PrayasAgrawal first give an answer.. :/ $\endgroup$ – user114592 Apr 19 '16 at 7:10
  • $\begingroup$ Now i have an answer. Sorry for late response. $\endgroup$ – Prayas Agrawal Apr 19 '16 at 7:13
  • $\begingroup$ To Muzamil, I have mentioned that $\vec{L}$ is not equal to $\vec{I_{axis}}\omega$ $\endgroup$ – Prayas Agrawal Apr 19 '16 at 7:16
  • $\begingroup$ Sorry for the wrong answer i completely misunderstood. @Prayas you are right:)The point must be ICR to apply Iw. $\endgroup$ – user114592 Apr 19 '16 at 7:18

The angular momentum will be different: however, you will be able to calculate it with the parallel axis theorem.

Measure the distance from your new axis to the center of mass and call it $d$. Your new rotational inertia $I$ can be calculated from the rotational inertia around the center of mass $I_{\mathrm{cm}}$ using the formula:

$$I = I_{\mathrm{cm}} + md^2$$

There is a derivation of this formula on Wikipedia.

Once you have found $I$, you can just find angular momentum using the same formula you were before: $L=I\omega$, where $\omega$ is the angular velocity about the new axis (the one that doesn't pass through the center of mass).

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    $\begingroup$ When referring to moment of inertia abtained by above theorem what will I take angular velocity? I.e its angular velocity about what axis? $\endgroup$ – Muzamil Sheikh Apr 19 '16 at 6:29
  • $\begingroup$ @MuzamilSheikh: The new axis (the one that doesn't go through the center of mass). $\endgroup$ – Deusovi Apr 19 '16 at 6:31
  • $\begingroup$ That is wrong. For correct expression read my answer. $\endgroup$ – Prayas Agrawal Apr 19 '16 at 7:06
  • $\begingroup$ @PrayasAgrawal What answer? $\endgroup$ – wgrenard Apr 19 '16 at 7:08
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    $\begingroup$ @PrayasAgrawal: You don't have an answer... $\endgroup$ – Deusovi Apr 19 '16 at 7:08

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