# Rotational Inertia of Two rods [closed]

In the figure, two particles, each with mass $= 0.82 \text { kg}$ are fastened to each other, and to a rotation axis at $O$, by two thin rods, each with length $d = 5.3 \text { cm}$ and mass $M = 1.0 \text { kg}$. The combination rotates around the rotation axis with angular speed $ω = 0.29 \text { rad/s}$. Measured about $O$, what is the combination's (a) rotational inertia and (b) kinetic energy?

The rotational inertia of a body about an axis to that of the same body when the axis is parallel is given by:

$$I=I_{1}+I_{2}+I_{3}+I_{4}$$

I_{1} is the moment of inertia of the rod that is closest to the axis of rotation. I_{2} is the moment of inertia of the particle that is nearest to the axis of rotation. I_{3} is the moment of inertia of the rod that is further away from the axis of rotation. I_{4} is the moment of inertia of the particle farthest away from the axis of rotation.

so that implies that:

$$I=I_{com}+mh^{2}$$

$$I=mk^{2}$$

$$I=\left[ I_{com}+M\left( h^{2}_{1}\right) \right] +m\left( k^{2}_{1}\right) ]+\left[ I_{com}+Mh^{2}_{2}\right] +m\left( k^{2}_{2}\right)$$

$$\dfrac {1}{12}Md^{2}\rightarrow I_{com}$$

$$\dfrac {d}{2}\rightarrow h_{1}$$

$$d\rightarrow K_{1}$$

$$\dfrac {3}{2}d\rightarrow h_{2}$$

$$2d\rightarrow k_{2}$$

$$I=\dfrac {8}{3}Md^{2}+5md^{2}$$

which yields I = 0.019

• Looks correct to me. Point masses, use $md^2 +m(2d)^2$. Treat the rods as one rod with mass $2M$ and length $2d$. So their m.o.I about O is $(1/3)(2M)(2d)^2$, which is what you get. Oct 25, 2014 at 23:59
• What is $h$? The figure shows $d$ for distances. Oct 26, 2014 at 5:07
• I corrected two typos in the left-hand bracket of your initial equation But this doesn't affect your result - which appears to be correct. Oct 26, 2014 at 15:32

Each rod has about its center of mass $I_{rod} = \frac{M}{12} d^2$. The center of mass of each rod is located $\frac{d}{2}$ and $\frac{3d}{2}$ from O respectively.
$$I_O = \left\{ I_{rod} + M \left(\frac{d}{2}\right)^2 \right\} + \left\{ m (d)^2 \right\} + \left\{ I_{rod} + M \left( \frac{3d}{2} \right)^2 \right\} + \left\{ m (2 d)^2 \right\}$$
Doing the simplifications yields $I_O = \frac{8}{3} M d^2 + 5 m d^2$, so you have the right formula.
• No, he had $I_{com}+m (h_1)^2$ instead of $I_{com} + M (h_1)^2$. Oct 26, 2014 at 13:44