Problem 2B-04*: The center of mass of a rod

(a) We have a uniform rod of length L. Show that the center of mass point of the rod is its center point.

(b) If the rod is held at one end and form an angle θ with a horizontal line, what is the torque around the held end point due to the gravitational pull exerted on this rod?

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Solution:

(a) Let us assume that the rod has a mass m. The center point of the rod is denoted as O. The rod is divided at O into two identical pieces. The right-hand side portion of the rod is dived into N identical pieces pieces each of that has a mass Δm = m/2N. The pieces are numbered from point O toward the right-hand end point of the rod as 1, 2, · · · , k, · ·, N. The rod is then folded along the center line and each of the piece in the right-had side is thus copied to the left-hand side of the rod. The pieces at the left-hand side are then labeled, starting from the one nearest to the center line, as -1, -2, · · ·, -k, · ·, -N. The set up is depicted in Fig. 1. The k-th piece and the (-k)-th piece are mirror emages with respect to the center line of the rod. The center of mass point of the k-th piece is denoted as Ck, and the center of mass point of the (-k)-th piece is denoted as C-k. It should be obvious that points C-k, O and Ck lie on one line. The poistion vector OCk is denoted as rk, and the position vector OC-k as r-k. Due to the symmetry with regard to the center line, we have

rk + r-k = 0 .

Thus

(r1 + r-1) + (r2 + r-2) + · · + (rN + r-N) = 0.                                 (1)

Since each piece has an idnetical mass Δm, Eq. (1) implies that Eq. (4) of Topic 2B-02 is satisfied. Therefore, the center point O is the center of mass point of this system of 2N pieces. Therefore, the center point O is the center of mass of the rod. It should be noted that N can be any positive number and the simplest case is apparently N = 1.

(b) Let the end of the rod held be denoted as point A as shown in Fig. 2. From the discussion that has led to Eq. (6) of Topic 2B-02, the torque around point A due to the gravitational pull on the rod can be considered to come from a point-like particle of mass m located at the center of mass point. The magnitude of the torque τ is (L/2)·m·g·cosθ. The torque τ originates from point A and points perpendicularly into the paper.