# “Sweet Spot” of Rod-Pendulum - Problem Clarification

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.

• if the rod were floating free, there would be a rotational moment depending on where you hit it. The rod would not rotate if you hit it at the center of mass and would have a forward motion without rotation. If it is being held in place by a pivot, the pivot point would "feel' an impulse in either direction, depending on where the rod is hit. – Peter R Apr 21 '16 at 0:46
• @PeterR Ah okay, so here it would the time-integral of the normal/restoring force that the pivot exerts on the rod, right? – Arturo don Juan Apr 21 '16 at 0:53
• @PeterR why don't you make that an answer.... – Floris Apr 21 '16 at 0:59
• I am not sure what you mean by a normal/restoring force, but depending on where it's hit, the rod would exert an impulsive force on the pivot and the pivot would exert an equal but opposite force. – Peter R Apr 21 '16 at 1:00

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).

As you have hypothesized in the comments, this will be a time integral of a force at the pivot point. As I read the problem, it would actually be the force exerted by the rod on the pivot that you're integrating, not the other way around.

More formally, the "impulse imparted on the pivot point" is a vector representing the total linear momentum transferred to the anchor from the rod during the collision.

(The "collision" here refers to an infinitessimal and not entirely realistic period of time in which energy and momentum redistribute themselves amongst the particle, the rod, and the anchor according to certain rules which are used for most basic thought experiments about rigid body collisions.)