Question

Distance of the center of mass of a solid uniform cone from its vertex is $z_0$ . If the radius of its base is $R$ and its height is h then $z_0$ is equal to :

• A. $\frac{h^2}{4R}$
• B. $\frac{3h}{4}$
• C. $\frac{5h}{8}$
• D. $\frac{3h^2}{8R}$

Answer 1

The correct option is B $\frac{3h}{4}$

Suppose the cylindrical symmetry of the problem to note that the center of mass must lie along the z axis (x = y = 0). The only issue is how high does it lie. If the uniform density of the cone is $\rho$ , then first compute the mass of the cone. If we slice the cone into circular disks of area $\pi r^2$ and height $dz$, the mass is given by the integral:

$M=\int \rho dV=\rho \underset{0}{\overset{h}{\int }}\pi r^{2}dz$

However, we know that the radius $r$ starts at a for $z=0$, and goes linearly to zero when $z=h$. This means that $r=a(1-z/h)$, so that:

$M=\rho \underset{0}{\overset{h}{\int }}\pi a^2(1-z/h{)}^{2}dz=\pi a^2\rho \underset{0}{\overset{h}{\int }}(1-2z/h+z^2/h^2)dz=1/3\pi a^{2}h\rho$

Now this simply indicates that the volume of the cone is given by

$V=\frac{1}{3}\pi a^{2}h$

To find the height of the center of mass, we then compute: $z_{cm}=\frac{1}{M}\int \rho zdV=\frac{\rho }{M}\underset{0}{\overset{h}{\int }}\pi r^{2}zdz=\frac{\pi a^2\rho }{M}\underset{0}{\overset{h}{\int }}(1-\frac{z}{h}{)}^{2}zdz$ $=\frac{\pi a^2\rho }{M}\underset{0}{\overset{h}{\int }}(z-\frac{2z^2}{h}+\frac{z^3}{h})dz=\frac{1}{12M}\pi a^2{h}^2\rho =\frac{1}{4}h$

As a result, the center of mass of the cone is along the symmetry axis, one quarter of the way up from the base to the tip and $\frac{3}{4}h$ from the tip.