Question

Spherical ball and solid cylinder have same mass and density. the moment of inertia will be greater for spherical ball or solid cylinder? Please answer me and explain the answer simply.

Answer 1

Answer on Question #37962, Physics, Mechanics | Kinematics | Dynamics

Let index one corresponds to sphere, and index two to cylinder. Hence, one has set of parameters $R_{1}, m, \rho$ for sphere and $R_{2}, h, m, \rho$ for cylinder ( $R$ is radius, $h$ is the height of cylinder, $m$ and $\rho$ are mass and density respectively).

Mass and density of cylinder are equal. Hence, $m = \frac{4}{3}\pi \rho R_1^3 = h\pi R_2^2\rho$ , which yields $R_{2}^{2} = \frac{4R_{1}^{3}}{3h}$ .

The moment of inertial of solid ball is $J_{1} = \frac{2}{5} mR_{1}^{2}$ and of cylinder $J_{2} = \frac{1}{2} mR_{2}^{2}$ . Hence,

$$\frac {J _ {1}}{J _ {2}} = \frac {4 R _ {1} ^ {2}}{5 R _ {2} ^ {2}} = \frac {3 h}{5 R _ {1}}.$$

Thus, which moment of inertia is higher depends on proportions of height of cylinder and radius of spherical ball (if their masses are equal):

$$J _ {1} > J _ {2} \Rightarrow h > \frac {5 R _ {1}}{3}$$ $$J _ {1} < J _ {2} \Rightarrow h < \frac {5 R _ {1}}{3}.$$