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

189.52 kg*m^2

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:

Parallel axis theorem for the rod gives:

`I_r=m_r\cdot((L)/(2)+(R)/(2))^2`

Paraller axis theorem for the spehere gives:

`I_s=(13)/(20)\cdot m_s\cdot R^2`

Therefore:

Replacing:

`\begin{gathered} I=(1)/(12)\cdot8\cdot6^2+8\cdot((6)/(2)+(1.5)/(2))^2+(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