A tiny rotating robotic arm weighs 0.5 N and has a mass radius of gyration of 2.0 cm. The mass moment of inertia $$(kg*cm^2)$$ is most nearly:

The mass radius of gyration is 2.0 cm.  We've been given the R in equation $$M*R^2 = I$$.

To find M, simply recognize that weight in newtons (N) is mass * gravity.

$$M ~~ (0.5 N) /(9.8 m/s^2) ~~ 0.051 kg$$

Therefore, the mass moment inertia $$I ~~ 0.051 * 2^2 ~~ 0.204 kg*cm^2$$

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The intuitive way to think about the mass radius of gyration is to imagine a rod rotating about its end. 

Note: Inertia of a particle $m$ a distance $r$ from an axis:  $$m*r^2$$

A rod is just a series of particles at further distances away from the axis of rotation.  We can assume that all these particles are the same mass and use the distributive property to extract m. 

$$I = m_1r_1^2 + m_2r_2^2 + ...+m_nr_n^2 => I = m*(r_1^2+r_2^2+... +r_n^2)$$

Based on this equation, we know that the total inertia of the rod is the mass of a particle multiplied by the sum of squared radii away from the axis of rotation.  Here's the trick:

Now imagine that a single large particle $M$ is floating some distance $R_g$ away from the axis.

$$M*R_g^2 = I = m*(r_1^2+r_2^2+... +r_n^2)$$

Apply a little trick: $$M/n = m$$

$$M*R_g^2 = M/n*(r_1^2+r_2^2+... +r_n^2)$$

$$R_g^2 = (r_1^2+r_2^2+... +r_n^2)/n$$

This shows that a single radius can be equated to all the radii of the particles that make up the rod.