Question

Determine the moment of inertia about the axis of the object shown in FIGURE P12.51.

Short answer

The total moment of inertia is $I_{Total}=\frac{M{L}^2}{12}+\frac{m_1{L}^2}{4}+\frac{m_2{L}^2}{16}$

Step-by-step solution

Given Information

The axis passes through the midpoint of the rod and is perpendicular to the rod.
Two balls of mass m₁and m₂ are connected as shown in the figure.
The mass of the rod is M and the length of the rod is L.

Explanation

The moment of inertia of a rod of mass M and length L, where it is about an axis passing through the mid point of the rod and perpendicular to the rod is given as

$I=\frac{M{L}^2}{12}........................................\left(1\right)$

As radius of the mass is not given lets consider them as point mass.

The moment of inertia of point mass is given by I=mr²

Now find the moment of inertia of both point mass

Mass-1

$I_1=m_1{\left(\frac{L}{2}\right)}^2=\frac{m_1{L}^2}{4}..............................\left(2\right)$

Mass-2

$I_1=m_2{\left(\frac{L}{4}\right)}^2=\frac{m_2{L}^2}{16}..............................\left(3\right)$

Add (1) , (2) and (3) to get total inertia.

$I_{Total}=\frac{M{L}^2}{12}+\frac{m_1{L}^2}{4}+\frac{m_2{L}^2}{16}$