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Angular impulse on a pivoted rod from two different points

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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.)

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?

asked Sep 25, 2017 in Physics Problems by samjoe (110 points)
edited Sep 27, 2017 by samjoe
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