Tricky hit

We wish to apply a force $F$ such that the acceleration of the bottom of the stick is zero. This acceleration is a combination of acceleration due to translation of the center of mass, and rotation about the center of mass of the stick: $a - \tfrac12\ell\alpha = 0.$ The linear acceleration is $a = F/m$ . For the rotational acceleration, we consider the torque $\tau = Fd$ with $d = \tfrac12\ell - h$ , and the rotational inertia $I = \tfrac1{12}m\ell^2$ . Thus $\alpha = \frac\tau I = \frac{F(\tfrac12\ell - h)}{\tfrac1{12}m\ell^2} = 12(\tfrac12 - h/\ell)\ \frac F {m\ell}.$ Substituting this into the first equation, we obtain $\frac F m - 6 \ell\left(\tfrac12 - \frac h \ell\right)\ \frac F{m\ell} = 0; \\ \frac F m\left(1 - 3 + 6 \frac h \ell\right) = 0 \\ 6 \frac h \ell = 2 \\ h = \boxed{\frac 13} \ell.$

We hit the stick with a horizontal force $F$ for a short time of $\Delta t$ . The momentum is $p= F \Delta t$ . The velocity of the center of mass will be $F\Delta t=p/m$ , horizontally forward. If we hit the rod at a distance $l/2$ from the top (i.e. we hit the center of mass) every point of the rod moves with this velocity.

Our goal is to create a "forward" rotation for the rod, so that the bottom moves backward and the top moves forward. More precisely, we want the bottom to move backward with a velocity of $v'=-v$ relative to the center of mass. We can achieve this if we hit the stick at a distance $h=al$ that is less than $l/2$ from the top.

The torque around the center of mass is $\tau=F (l/2-h)=F l(1/2-a)$ . The corresponding change in angular momentum is $F l(12/-a) \Delta t$ . The moment of inertia of the rod is $\frac{1}{12} ml^2$ , and the angular momentum after the kick is $\frac{1}{12} ml^2 \omega = l(1/2-a) F \Delta t$ . Since $F\Delta t = mv$ we get

$\frac{1}{12} ml^2 \omega = l(\frac{1}{2}-a) mv$ .

Due to the rotational motion the bottom of the stick will move backwards with a velocity of $v'=-\omega \frac{l}{2}$ relative to the center of mass. Our goal is the have no horizontal velocity component for the bottom of the stick. The is achieved if $\omega \frac {l}{2} = v$ . Inserting that to the equation above yields $a=1/3=0.333$

Notes

• We assumed that the force $F$ was constant during the time $\Delta t$ . The result is the same if the force is time dependent, but we have to use $\int F dt$ instead of $F \Delta t$ . In either case the time of collision must be much less than $1/\omega$ (i.e. the stick does not move significantly during the collision).

• If you want to try this, mark up a yardstick for the proper hitting point. Place a little bit of play-dough to the bottom of the cup and press the yardstick into it so that it stands vertical. Hit with a shorter, more massive stick.

• As the center of mass moves horizontally (at least initially) and the end point of the stick makes a circular potion around the center, the orbit of the end point will be a cycloid. The figure below illustrates the motion of the center of mass (blue) and the end point (yellow). The tip of the arrow represents the end point of the stick. Right after the hitting of the stick the end point moves vertically upwards.