Problem 2B-11**: Center of mass of a spherical cone

Find the center of mass of a uniform spherical cone of height h. The bottom of the cone has a radius d as shown in Fig. 1.






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Solution:

Fig. 2 shows a side view of the cone. The cone is cut into N parallel layers with an equal thickness Δh = h/N. Each layer is approximated by a uniform circular disk. The radius of the k-th layer is denoted as rk and the center of mass as Ck. The density of the material that makes up the uniform cone is denoted as ρ. The volume of the k-th disk, ΔSk, is

                ΔSk = πrk2·Δh = πrk2·h/N.

The mass of the k-th disk, mk, is

                Δmk = ρ·ΔSk = (πρh/N)· rk2.

The radius rk can be computed from the configuration of Fig. 3 as

                rk = k·Δh·d/h = k·d/N.

Thus

                Δmk = k2·(πd2ρh/N3).                                 (1)



The problem to find the center of mass point C(N) of the N-circular disk system is now reduced to the problem of finding the center of mass of N mass points located at C1, C2, · · ·, CN, and with masses m1, m2, · · ·, mN respectively as shown in Fig. 4. The position vector ACk is denoted as Rk. It is obvious that every Rk originates from point A, lies on line AO and points toward point O. We define a unit length position vector j that originates from point A, lies on line AO and points toward point O. The position vector Rk can be written as

                Rk = Rk·j.

Since

                Rk = ACk = k·Δh - Δh/2 = (k - 1/2)·(h/N),

we have

                Rk = (k - 1/2)·(h/N)·j.                                 (2)

The position vector AC(N) is denoted as R(N). Eq. (5) of Topic 2B-02 says that

                R(N) = (m1R1 + m2R2 + ··· + mNRN) / (m1 + m2 + ··· + mN).

Using Eq. (1) and Eq. (2), we have

                m1R1 + m2R2 + ··· + mNRN = {Σ(k=1 to N)m k·(k - 1/2)}·(h/N)·j
                                = {Σ(k=1 to N) k2·(k - 1/2)}· (πd2ρh2/ N3j,

and

                m1 + m2 + ··· + mN = {Σ(k=1 to N)k2}· (πd2ρh/N3).

Thus

                R(N) = [{Σ(k=1 to N)k2·(k - 1/2)} / {Σ(k=1 to N)k2}]·(h/N)·j.

Since

                Σ(k=1 to N)k2 = N·(N + 1)·(2N + 1)/6,

                Σ(k=1 to N)k3 = N2·(N + 1)2/4,

and

                Σ(k=1 to N)k2(k - 1/2) = {Σ(k=1 to N)k3} - (1/2)·{Σ(k=1 to N)k2},

we have

                {Σ(k=1 to N)k2·(k - 1/2)} / {Σ(k=1 to N)k2} = [{Σ(k=1 to N)k3} / {Σ(k=1 to N)k2}] - (1/2)
                                                                          =(1/2)·[{3N·(N + 1)/(2N + 1)} - 1].

Thus

                R(N) = (1/2)·[{3N·(N + 1)/(2N + 1)} - 1]·(h/2)·j.

As N → ∞, C(N) → C, where C is the center of mass of the actual cone, and R(N) → R, where R is the position vector AC. Therefore,

                R = limN→∞R(N) = (3/4)·h·j.

In summary the center of mass of a cone of height h is located on the center line AO and is a distance 3h/4 away from point A.