My actual question is that does the point about which angular momentum relation is calculated need to be inertial? I mean do we have to be in the reference frame of a point to write angular momentum about that point? This is illustrated in the following question, particularly in Method 2

Q. Suppose we have a rod pivoted at the top. A linear impulse $J$ is given to rod at point B in horizontal direction. We need to calculate the linear impulse $J'$ provided by pivot. (Length of rod is $L$, mass is $m$, moment of inertia is $I = \frac{mL^2}{3}$ about pivot. Also consider that rod gets an angular velocity $\omega$ just after the impulse acts.)

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I did this problem using two different approaches:

Method 1: Angular momentum and impulse relation about point $A$ (pivot).

Total angular implulse (external) is $JR$ about point $A$, or the pivot. This must equal change in angular momentum, ie, $I \omega$. So we obtain $$J = \dfrac{I \omega}{ L} = \dfrac{m\omega L}{3}$$ Also we know that net linear impulse on rod is equal to change in its linear momentum. we obtain $$J - J' = \dfrac{m\omega L}{2}$$

From these two equations we obtain $$J' = \dfrac{-m\omega L}{6}$$

Method 2: Angular momentum about point B.

I started reasoning as follows: Since an impulse acts on rod at point $B$, then if we take reference point to be $B$, then we apply a pseudo force from center of mass $C$, or an impulse $J$ from $C$ in opposite direction. Now we write angular momentum - angular impulse relation about $B$.

$$\dfrac{JL}{2} + J'L = I \omega = \dfrac{mL^2\omega}{3}$$

using the value of J from above, we get:

$$J' = \dfrac{m\omega L}{6}$$

Which is clearly different from result of method 1.

To me, method 1 seems fine but there seems to be a major conceptual error in method 2. Can you please explain more about angular momentum in this situation?


Pseudo-forces are applied in non-inertial reference frames. You have not been clear about what reference frame you are using. A reference point is not a reference frame. Point B is - within the same instant - at rest (before the impulse), accelerating (during the impulse), and moving with constant instantaneous velocity (after the impulse). These are 3 different reference frames.

If you are using a frame in which B is moving with constant horizontal velocity - eg immediately before or immediately after the impulse is applied - then this is an inertial frame so it is wrong to apply a pseudo-force. If you are using a non-inertial frame you must be clear whether you the frame accelerates in a straight line or moves with constant speed while it rotates about point A.

  • $\begingroup$ So is method 1 correct? In method 2, if we consider the pseudo force or not, the answer doesn't match with that of method 1. Please tell me which is more appropriate here? Going by your answer, I understand that it is not possible for us to use method 2, because it has some variable centripetal acceleration and some tangential acceleration too. Is there any way to make method 2 work? $\endgroup$ – samjoe Sep 27 '17 at 4:48
  • $\begingroup$ Method 1 is correct. Point A remains at rest throughout the motion, so there is no difficulty in using a frame in which A is at rest. Method 2 is incorrect. Yes I think it could be made to work, but you have to identify what frame of reference you are attempting to use. $\endgroup$ – sammy gerbil Sep 27 '17 at 11:54
  • $\begingroup$ Thanks a lot! This is really useful. If possible, please let me know how I can make method 2 work. $\endgroup$ – samjoe Sep 27 '17 at 12:07
  • $\begingroup$ Can you tell me the reason for the impulse by the pivot..why is it present and even at the centre also as shown in the diagram above??Also How can we use B as a reference point when it has a torque acting along it's axis $\endgroup$ – Hydrous Caperilla Apr 3 '18 at 5:26
  • $\begingroup$ @HydrousCaperilla If pivot A were not there the impulse at B sets the rod rotating anti-clockwise about its centre C, while C moves to the right. The pivot at A prevents the rod from moving at A, so it exerts a reaction impulse $J'$ on the rod. This impulse should be to the right also, not to the left as in the diagram. ... The impulse at B is equivalent to an impulse (to the right) at C and an anti-clockwise torque about C. samjoe's diagram is misleading. $\endgroup$ – sammy gerbil Apr 3 '18 at 9:42

There is no angular impulse by the pivot when A is your pivot point. This is just not possible because there's no lever arm. Also, your units do not match. How can angular impulse and linear impulse have the same units?. The dimensional analysis just doesn't work and your solution is completely flawed, even without dealing with the reference frames.

  • $\begingroup$ Exactly which line are you referring to? $\endgroup$ – samjoe Sep 21 '17 at 4:36
  • $\begingroup$ your entire question, especially method one and method two. $\endgroup$ – Will-i-am Guo Sep 22 '17 at 5:18
  • $\begingroup$ I have taken linear impulse as $J$. So consider accordingly. $\endgroup$ – samjoe Sep 22 '17 at 8:43

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