# Center of percussion

(Redirected from Center of oscillation)

The center of percussion is the point on an extended massive object attached to a pivot where a perpendicular impact will produce no reactive shock at the pivot. Translational and rotational motions cancel at the pivot when an impulsive blow is struck at the center of percussion. The center of percussion is often discussed in the context of a bat, racquet, door, sword or other extended object held at one end.

The same point is called the center of oscillation for the object suspended from the pivot as a pendulum, meaning that a simple pendulum with all its mass concentrated at that point will have the same period of oscillation as the compound pendulum. In sports, the center of percussion of a bat or racquet is related to the so-called "sweet spot", but the latter is also related to vibrational bending of the object.

## Explanation Effects of a blow on a hanging beam. CP is the Center of Percussion, and CM is the Center of Mass of the beam.

Imagine a rigid beam suspended from a wire by a fixture that can slide freely along the wire at point P, as shown in the Figure. An impulsive blow is applied from the left. If it is below the center of mass (CM) it will cause the beam to rotate counterclockwise around the CM and also cause the CM to move to the right. The center of percussion (CP) is below the CM. If the blow falls above the CP, the rightward translational motion will be bigger than the leftward rotational motion at P, causing the net initial motion of the fixture to be rightward. If the blow falls below the CP the opposite will occur, rotational motion at P will be larger than translational motion and the fixture will move initially leftward. Only if the blow falls exactly on the CP will the two components of motion cancel out to produce zero net initial movement at point P.

When the sliding fixture is replaced with a pivot that cannot move left or right, an impulsive blow anywhere but at the CP results in an initial reactive force at the pivot.

## Calculating the center of percussion

For a free, rigid beam, an impulse $Fdt$ applied at right angle at a distance $b$ from the center of mass (CM) will result in the CM changing velocity $dv_{cm}$ according to the relation:

$F=M{\frac {dv_{cm}}{dt}},$ where $M$ is the mass of the beam. Similarly, the torque about the CM will change the angular velocity $\omega$ according to:

$Fb=I{\frac {d\omega }{dt}},$ where $I$ is the moment of inertia around the CM.

For any point P a distance $p$ on the opposite side of the CM from the point of impact, the change in velocity of point P is

$dv_{net}=dv_{cm}-pd\omega \,$ where $p$ is the distance of P from the CM. Hence the acceleration at P due to the impulsive blow is:

${\frac {dv_{net}}{dt}}=\left({\frac {1}{M}}-{\frac {pb}{I}}\right)F.$ When this acceleration is zero, $b$ defines the center of percussion. Therefore, the CP distance, $b$ , from the CM, is given by

$b={\frac {I}{pM}}.$ Note that P, the rotation axis, need not be at the end of the beam, but can be chosen at any distance $p$ .

Length $b+p$ also defines the center of oscillation of a physical pendulum, that is, the position of the mass of a simple pendulum that has the same period as the physical pendulum.

## Center of percussion of a uniform beam

For the special case of a beam of uniform density of length $L$ , the moment of inertia around the CM is:

$I={\frac {1}{12}}ML^{2}$ (see moment of inertia for proof),

and for rotation about a pivot at the end,

$p=L/2$ .

$b={\frac {L^{2}}{12p}}={\frac {1}{6}}L$ .
It follows that the CP is 2/3 of the length of the uniform beam $L$ from the pivoted end.