Problem 2B-06*: Centers of mass of a ring and a circular disk
(a) Show that the center of mass of a uniform ring is its center.
(b) Show that the center of mass of a uniform circular disk is its center.
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(a) Let us cut the ring to 2N identical pieces with a mass Δm for each piece.
Take one of such pieces and label it as the k-th piece. We denote the center of mass
of this piece as Ck, and the center of the ring as O.
The position vector OCk is denoted as
rk. In the symmetric position to the k-th piece
with respect to the center point O, there is an identical piece. We denote this
piece as the (-k)-th piece and its center of mass as C-k.
The position vector OC-k is denoted as
r-k. The configurations are depicted in Fig. 1.
From the symmetric consideration it should be obvious that we have
rk + r-k = 0.
We label the pieces from the k-th one clockwise as k+1, k+2, · · ·, N, and the pieces from the k-th one counter-clockwise as k-1, k-2, · · ·, 1. Similary the pieces from the (-k)-th one clockwise as -(k-1), -(k-2), · · ·, -N, and the pieces from the (-k)-th one counter-clockwise as -(k-1), -(k-2), · · ·, -1. We have
(r1 + r-1) + (r2 + r-2) + · · · + (rN + r-N) = 0 . (1)
Since each piece has an identical mass Δm, Eq. (1) means that Eq. (4) of Topic 2B-02 is satisfied. Therefore, the center point O is the center of mass of the ring. It should be noted that in this proof N can be any positive number. Thus the simplest case is to take N = 1 and just consider two half rings.
(b) A uniform circular disk can be considered as made of a finite number, say N, uniform cocentric rings as shown in Fig. 2. In part (a) we have shown already that the center mass of each of such rings is the center of the disk, point O. If we denote the position vector from O to the center of mass of the k-th ring, also point O, as rk, it is obvious that rk = 0. Thus Eq. (4) of Topic 2B-02 is automatically satisfied. Therefore, O is the center of mass of the circular disk.