# 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
edited Sep 27, 2017
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