I came across this problem in a book (shortened for brevity):
Consider a rod of mass $m$ pivoted about one end, with the other end to rotate. Let the center of mass be a distance $a$ from the pivot point $I$ be the moment of inertia of the rod about an axis which we will consider rotations in. A particle comes in and hits the rod at a distance $b$ below the pivot point, imparting an impulse $F\Delta t=\xi$ on the rod. (a) Find the linear and angular momentum of the rod right after the time $\Delta t$, and (b) Calculate the impulse imparted on the pivot point.
My problem is with (b). What the does "impulse imparted on the pivot point" even mean? I would think the pivot point is fixed, so it should have experienced no net impulse, but that's incorrect.
Answer
In order to maintain the constraint of the pivot during the impact, a reaction impulse is needed. See the figure below for what I mean.
At the center of mass the velocity is $v = a\,\omega$. This is a result of the two impulses $$(F-R) \Delta t = m\, a\, \omega$$
If the angular velocity is $\omega$ then the net impulsive moments at the center of mass are
$$ (b F + a R) \Delta t = I \omega $$
These two equations are solved for the unknown reaction $R$ and motion of the rod $\omega$.
$$\begin{aligned} R \Delta t & = \frac{I-m\,ab}{I+m a^2} F \Delta t \\ \omega & = \frac{a+b}{I+m a^2} F \Delta t \end{aligned}$$
Only when $b=\frac{I}{m\,a}$ the pivot reaction is zero. That is considered the instant axis of percussion of the rod about the pivot (sweet spot).
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