How do I find the position of the centre of percussion of a sector of a circle, axis in the plane of the sector, perpendicular to its symmetrical radius, and passing though the center of the circle?

The center of percussion of a sector of a circle is the point on the symmetrical radius where a perpendicular impact will produce no reaction at the axis of rotation.

Let's consider a sector of a circle with radius r and central angle θ. The center of the circle is denoted by O, the center of the sector is denoted by A, and the symmetrical radius is denoted by AB. The axis of rotation is perpendicular to AB and passes through the center of the circle O.

To find the position of the center of percussion C, we can use the following steps:

1. Draw a line through A perpendicular to AB, and let it intersect the axis of rotation at point P.

2. Draw a line through C perpendicular to AB, and let it intersect the line AP at point Q.

3. The distance AQ is the position of the center of percussion C.

The proof of this can be shown as follows:

Let P be the point of impact and R be the point where the reaction force acts on the axis of rotation. When a force F is applied at P perpendicular to the sector, it will produce a reaction force -F at R in the opposite direction. The moment of the force F about the axis of rotation is F * OP, and the moment of the reaction force -F about the axis of rotation is -F * OR. In order for there to be no reaction force at the axis of rotation, the moments of the force and the reaction force must be equal and opposite, i.e. F * OP = F * OR, which implies that OP = OR.

Let Q be the center of percussion. The moment of the force F about Q is F * AQ, and the moment of the reaction force -F about Q is -F * QB. Since the point Q is chosen such that there is no reaction force at the axis of rotation, the moments of the force and the reaction force about Q must also be equal and opposite, i.e. F * AQ = F * QB, which implies that AQ = QB. Therefore, AQ = QB = OP = OR, which means that Q is the center of percussion.

To find the distance AQ, we can use the following formula:

AQ = AB * cot(θ/2)

where cot is the cotangent function. This formula can be derived by using the properties of right triangles and the fact that the tangent of half the central angle is equal to the ratio of the opposite side (AB) to the adjacent side (OA).

Therefore, the position of the center of percussion C is AQ = AB * cot(θ/2).