BAT LAB: The Sweet Spot
Dynamics of Machines
Given the center of rotation, find the sweet spot of a baseball bat. In the process, find:
Also, compare theoretical predictions of these values to experimental data.
If you've ever played baseball or tennis, you've probably heard of the term "sweet spot." You also probably know that this is where you want to hit the ball to crank it out of the park. If you hit a ball too close to your grip (ball A in the figure below), it stings the palm of your top hand like a motha'. If you hit the ball too far up the head (ball B), it might sting (but not as bad) the palm of your bottom hand. If you hit it in the sweet spot (ball C), oh man, it feels like butta'. Why is this the case?
When you strike an object at its center of gravity (CG), the object will accelerate in the direction of the applied force in pure translation. If you strike an object off its center of gravity, it will rotate about its CG in addition to the translation. When ball A (in the above figure) impacts the bat, the CG will accelerate down as well rotate counterclockwise (as you can see from inspection). Thus, at the grip point, the bat is accelerating down due to translation and accelerating down due to rotation (which explains why it hurts!). When ball B impacts the bat, on the other hand, the CG will accelerate down but rotate clockwise. At the grip, the bat will accelerate down due to translation but up due to rotation. The upward acceleration from rotation will be greater than the downward acceleration. As you can imagine, there is some point where you could impact the bat such that the upward acceleration and downward acceleration cancel each other out. That point is the sweet spot or, more technically, the center of percussion. The grip, in this case, is the center of rotation--the point where the net acceleration is zero. It is important to note that there is not a single center of percussion nor a single center of rotation; they come in pairs. In other words, the sweet spot changes depending on where you grip the bat.
Now how would you go about determining the center of rotation/center of percussion pair? Finding the center of gravity is a good place to start. The center of gravity depends on the mass distribution in the bat. The mass distribution can be characterized by the mass moment (or first moment of mass) about the x, y and z axes. About the x-axis (the only axis about which the bat is not symmetric), the mass moment can be calculated by:
The mass moment, in words, is a differential mass element times its distance from the axis in question. The above equation is handy if you have a function describing the geometry of the bat, but that's probably not the case. Another method is to break the bat into simple geometrical shapes for which we know the the center of gravity. For example, you could model the bat as two cylinders; a small-diameter long cylinder for the shaft and a larger-diameter short one for the the head. Given that the mass moment is mass times distance, the equation can be simplified as:
where mi is the mass of the ith cylinder and xi is the distance from the origin of the x-axis to the CG of the ith cylinder. This says that the mass moment of the sum of the components is equal to the mass moment of the bat as a whole. The distance d is the distance to the center of mass (or center of gravity) of the bat as a whole. A more general solution for the center of gravity, with any number of shapes representing the bat, is given by:
where CG is the distance from the origin (from where each of the d's were measured) to the center of gravity. The physical meaning of the CG is that this is the point where the masses on either side of it are equal; thus, you could put your finger beneath this point and the bat would balance.
Just as mass is a measure of resistance to translational acceleration (F=ma), moment of inertia (or second moment of mass) is a measure of resistance to rotational acceleration (T=Iα). The location of the center of percussion depends on the bat's moment of inertia, which can be calculated by:
But just as the first moment of mass can be broken up into geometric shapes, so can the second moment of mass. One can look up values for moments of inertia for various volumes and then apply the following equation to obtain the moment of inertia of the bat as a whole about a given axis:
This is the parallel axis theorem which says that the moment of inertia can be given about any axis (IZZ) if you know the moment of inertia about a parallel axis through the CG (ICG) that is a distance d away. So, after breaking up the bat into various simple volumes (cylinders, spheres, frustums, etc), sum the moment of inertia for each volume about a known axis (e.g., an axis that goes through the CG) plus that volume's mass times its distance, squared, from the axis for which you want to know the moment of inertia. You can also sum negative volumes. For example, to find the moment of inertia of a truncated cone, subtract the moment of inertia of the negative volume (cone tip) from the moment of inertia of the whole cone.
It is also possible to determine moment of inertia experimentally. If you swing the bat like a pendulum (as shown in the figure below), its frequency is a function of its moment of inertia about the rotational point (IO).
Summing the torques (moments) about O and using the small angle approximation (sin(θ) ≈ θ) yields:
where θ is the angle off vertical and LCG is the distance from O to the center of gravity. If you swing the bat from its handle, LCG is simply the distance you measured/calculated from the end of the bat (plus the distance to the eyehook). If you use a setup as shown in the figure above, LCG is given by (using the Law of Cosines):
where L1 and L2 are the lengths of the strings attached to the handle and head ends of the bat, respectively. The natural frequency of this pendulum is:
where Ωo is in rad/sec. So, if you measure the time it takes for the bat to swing back and forth once (its period), the natural frequency can be determined. The moment of inertia about O is then simply:
Using the parallel axis theorem, the moment of inertia about any point on the bat (e.g., the center of gravity or grip point/center of rotation) can be calculated if you know the distance from O to that point. Once you know the moment of inertia about a given axis, IZZ, it is often useful to know the radius of gyration--the distance from the ZZ axis such that the entire mass of the bat could be concentrated and still have the same moment of inertia. Convenient when creating a simple model (i.e., point mass) of the system, the radius of gyration is given by the simple equation:
Finally, we are ready to find the center of rotation/center of percussion pair. If the center of rotation is a distance x from the CG, then the distance from the CG to the center of rotation is given by (see the text for derivation):
where the radius of gyration, k, is calculated with respect to the IZZ, axis through the CG. The negative sign implies that the center of rotation and center of percussion lie on opposite sides of the CG.
By now, you should see that the the center of percussion, or sweet spot, is completely dependent on where you hold the bat. For example, if you "choke up" on the bat (move your grip up), the sweet spot will be different than your normal grip.
MATERIAL AND METHODS:
This lab requires:
|MOMENT OF INERTIA||
|RADIUS OF GYRATION||
|CENTER OF PERCUSSION|| |
|MOMENT OF INERTIA|| |
|RADIUS OF GYRATION|| |
|CENTER OF PERCUSSION|| |
THE "REAL" SWEET SPOT
|CENTER OF PERCUSSION||
Alternative procedure (if jig available):