Move the metal ball up or down the string until its period of oscillation matches that of the baseball bat. When the periods of oscillation match, the position of the ball corresponds to the position of the center of oscillation of the bat, also known as the center of percussion. When you strike the bat at this point with the wooden dowel (from the front), there is no displacement of the pivot point, and the bat undergoes pure rotation. Also, when you strike the bat at the center of percussion, the sound is noticeably different from the sound produced when you hit the bat at any other point.
This apparatus consists of a miniature baseball bat suspended by a pivot rod that rests on a yoke. The rod sits in notches in the yoke, and has two notches that keep it from sliding sideways on the yoke. Three holes drilled in the handle of the bat allow you to choose three points at which to set the pivot. The baseball bat is free to swing about the pivot in a plane perpendicular to the front of the table, and if you displace it either to the front or to the back and release it, it oscillates. It is a physical pendulum, an extended (rigid) object suspended from a point either at its edge or within its outline, as opposed to a simple pendulum, in which the mass is concentrated at a point, and the object is suspended by a massless string of a particular length. If we suspend a physical pendulum from any point, it comes to rest in the position in which its center of mass sits below that point on a vertical line. (See demonstration 24.33 -- California board.) We will call the suspension point, about which the pendulum can pivot, P, the center of mass C, and the distance between them d.
If we displace the pendulum from its equilibrium position by an angle θ, gravity exerts a torque at C about P to bring the pendulum back to its equilibrium position. This torque is
τ = −Mgd sin θ
where M is the mass of the pendulum. As we did in the analysis for demonstration 40.21 -- Pendulums of different lengths and masses, we may simplify things by limiting the displacement to small angles, in which case θ ≈ sin θ, and the torque becomes
τ = −Mgd θ
the angular acceleration due to this torque is
τ = I(d2θ/dt2) = Iα
So we have
(d2θ/dt2) + (Mgd/I) θ = 0
This is the same differential equation that describes the motion for demonstration 40.21, mentioned above, and for demonstration 40.12 -- Mass-springs with different spring constants and masses. This gives the frequency of oscillation as ω = √(Mgd/I). The period is 2π/ω, or
T = 2π√(I/Mgd)
As noted above, this holds only for oscillations of small amplitude. It is interesting to note what happens when we use this expression for a point mass m suspended from a massless string of length l. The moment of inertia is I = ml2, M = m, and d = l, which gives
T = 2π√(I/Mgd) = 2π√(l/g)
which is the period we found for the simple pendulum in demonstration 40.21. This equation also shows us what length a simple pendulum must be for it to have the same period as a physical pendulum. If we take the middle and right terms above, cancel the factors of 2π and √g, and square, we get
l = (I/Md)
The physical pendulum behaves as if its mass were concentrated at a point located at a distance of l = (I/Md) from the pivot point. This point, which we may call O, is called the center of oscillation of the physical pendulum. There are two interesting properties associated with this point.
To show the first property, we will find the center of oscillation of a thin rod of length l suspended from one end. The distance between the pivot and the center of mass, d, equals l/2. The moment of inertia for the rod rotated at its end is I = Ml2/3, so T = 2π√[I/(Mgd)] = 2π√[(Ml2/3)/Mg(l/2)] = 2π√[(2/3)(l/g)]. The length of a simple pendulum that has this period is I/Md, which equals (2/3)l. The center of oscillation O of the rod, then, is at a distance (2/3)l from the pivot P.
Now let us exchange the positions of P and O. The pivot, P, is now at a distance l/6 from the center of mass. The moment of inertia about the center of mass is Ml2/12. The moment of inertia about P is thus Ml2/12 + Ml2/36 = (1/9)Ml2. If we insert d = l/6 and I = (1/9)Ml2 into the expression for T, we get
T = 2π√[(1/9)Ml2/Mg(l/6)] = 2π√(2/3)(l/g)
The length of the simple pendulum that has this period is (2/3)l, the same length that we found for the rod suspended at the end. If we suspend the rod where the center of oscillation was, the original pivot point is now the center of oscillation; the rod has the same period of oscillation whether we suspend it from point P or from point O. For this reason, the pivot point and center of oscillation are said to be conjugate to each other. (We could have chosen point P to be some distance from the end of the rod. For example, if we put P (1/6)l from the end of the rod, this puts it (1/3)l from C, makes I = (7/36)Ml2, and puts O at a distance of (7/12)l from P. Exchanging O and P then puts P (3/12)l from C, makes I = (21/144)Ml2, and again puts O at a distance of (7/12)l from P.)
Second, we consider what happens when we deliver an impulsive force to the bat by striking it at various points. If we strike the bat at its center of mass, the impulse exerts no torque on the bat; it imparts only translational motion to it. This generates a backward horizontal force at the pivot, displacing it backwards (toward the table). An interesting thing happens, however, if we strike the bat at its center of oscillation. To illustrate this, we return to the example of the thin rod. We again have the rod suspended from its end, and imagine that we strike it at O, the center of oscillation. This imparts both a translational motion and a rotational motion to the rod. If we strike the rod with force of magnitude F, in translating the rod this causes the pivot P to move with an acceleration a = F/M in the direction of our stroke. It also rotates the rod about the center of mass C, with the pivot moving in the opposite direction to our stroke. The rotational acceleration is α = τ/I = (F)(l/6)/(Ml2/12) = 2F/Ml. The linear acceleration of the pivot is a = α(l/2) = (2F/Ml)(l/2) = F/M, the same force as for the translation from the stroke, but in the opposite direction. These two forces thus cancel, and there is no net horizontal force on the pivot; it does not move. (Again, we could have chosen P to be some distance from the end of the rod, and the result would have been the same.) If we strike the rod anywhere else besides the center of oscillation, these forces either do not balance or are in the same direction, and there is a force at the pivot that would displace it horizontally.
Because an impulse delivered at the center of oscillation produces no displacement force at the pivot, this point is also called the center of percussion, hence the title of this demonstration.
As shown above, we can find the center of oscillation of our baseball bat by finding the length of a simple pendulum whose period of oscillation matches that of the bat. The ball hanging beside the bat provides an easy way to do this. The string is looped through the ball so that you can slide it up or down as needed, but once you set it, it will stay in place. Set the bat and ball oscillating. If the periods do not match, adjust the ball as necessary until they match. When the periods match, the height of the ball indicates the location of the center of oscillation of the bat. The apparatus is mounted on a 1/2-inch-diameter aluminum rod, because it is flexible enough to allow displacement of the pivot if a horizontal force acts there.
Once you have found the center of percusion, when you strike the bat there with the wooden dowel, the bat swings, but the pivot does not move (or the movement is very slight). When you hit the bat at the center of percussion, the sound produced is sharp and resonant.
If you hit the bat between the center of oscillation and the center of mass, the rotation no longer balances the translational impulse, and the pivot moves backward (toward the table). When you hit the bat above the center of mass, the translation and the rotation are in the same direction, and the pivot moves backward. The further from the center of oscillation you hit the bat, the greater the rearward displacement of the pivot.
If you hit the bat below the center of oscillation, the rotation now exceeds the translation, and the pivot moves forward (away from the table). Again, the further from the center of oscillation you hit the bat, the greater the forward displacement of the pivot. In comparison to the sound produced when you hit the bat at the center of oscillation, the sound produced when you hit the bat anywhere else is hollow and dull.
As noted above, you have a choice of three holes through which to place the pivot of the bat.
The center of percussion is of particular interest to baseball players. When a baseball player hits a baseball, if the ball makes contact with the bat at its center of percussion, there is no displacement at the handle. If the ball makes contact anywhere else on the bat, however, a force is generated on the handle, which hurts the baseball player’s hands and wrists. This is probably why the object chosen as the physical pendulum for this demonstration is a baseball bat.
References:
1) Resnick, Robert and Halliday, David. Physics, Part One, Third Edition (New York: John Wiley and Sons, 1977), pp. 240-242, 313-315.
2) Sears, Francis Weston and Zemansky, Mark W. College Physics, Third Edition (Reading, Massachusetts: Addison-Wesley Publishing Company, Inc., 1960), pp.232-234.