Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 8 kg and length L = 6 m to a uniform sphere with mass ms = 36.25 kg and radius R = 1.5m.What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)

Answer 1

ANSWER:

189.52 kg*m^2

STEP-BY-STEP EXPLANATION:

Given:

mr = 8 kg

ms = 36.25 kg

L = 6 m

R = 1.5 m

Moment of inertia of rod about its center of mass:

$I_{\operatorname{cm}}=\frac{1}{12}\cdot m_r\cdot L^2$

Parallel axis theorem for the rod gives:

$I_r=m_r\cdot(\frac{L}{2}+\frac{R}{2})^2$

Paraller axis theorem for the spehere gives:

$I_s=\frac{13}{20}\cdot m_s\cdot R^2$

Therefore:

$\begin{gathered} I=I_{\operatorname{cm}}+I_r+I_s \\ I=\frac{1}{12}m_rL^2+m_r(\frac{L}{2}+\frac{R}{2})^2+\frac{13}{20}m_sR^2 \end{gathered}$

Replacing:

$\begin{gathered} I=\frac{1}{12}\cdot8\cdot6^2+8\cdot(\frac{6}{2}+\frac{1.5}{2})^2+\frac{13}{20}\cdot36.25\cdot1.5^2 \\ I=189.52\text{ kg}\cdot m^2 \end{gathered}$

The moment of inertia is 189.52 kg*m^2