Step 1 We are provided with the following data:
The mass of circular disk A is ${m_A} = 8{\rm{ kg}}$.
The mass of circular disk B is ${m_B} = 2{\rm{ kg}}$.
The mass of slender rod is ${m_r} = 4{\rm{ kg}}$.
The length of slender rod is ${l_r} = 1.5{\rm{ m}}$.
The distance between the point O and the disk A is ${x_A} = 1{\rm{ m}}$.
The distance between the point O and the disk B is ${x_B} = 0.5{\rm{ m}}$.
We are required to determine the radius of gyration of the pendulum about an axis perpendicular to the page passing through the point O.
Step 2 The diagram can be represented as,
Here, ${r_A},{\rm{ }}{r_B}$ represents the radius of mass A and mass B, ${m_A},{m_B},{m_r}$ represents the mass of the sphere A, B and the rod.
Step 3 To find the moment of inertia of disk A, we will use parallel axis theorem,
\[I{ _A} = \frac{{{m_A} \times r_A^2}}{4} + {m_A}{\left( {{r_A} + {x_A}} \right)^2}\]
On plugging the values in the above relation, we get,
\[\begin{array}{c} {I_A} = \frac{{8{\rm{ kg}} \times {{\left( {0.2{\rm{ m}}} \right)}^2}}}{4} + 8{\rm{ kg}} \times {\left( {0.2{\rm{ m}} + 1{\rm{ m}}} \right)^2}\\ {I_A} = \frac{{8{\rm{ kg}} \times 0.04{\rm{ }}{{\rm{m}}^2}}}{4} + 8{\rm{ kg}} \times {\left( {1.2{\rm{ m}}} \right)^2}\\ {I_A} = \frac{{8{\rm{ kg}} \times 0.04{\rm{ }}{{\rm{m}}^2}}}{4} + 8{\rm{ kg}} \times 1.44\;{{\rm{m}}^2}\\ {I_A} = 0.08{\rm{ kg}} \cdot {{\rm{m}}^2} + 11.52\;{\rm{kg}} \cdot {{\rm{m}}^2}\\ {I_A} = 11.6\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{array}\]
Step 4 To find the moment of inertia of disk B, we will use parallel axis theorem,
\[I{ _B} = \frac{{{m_B} \times r_B^2}}{4} + {m_B}{\left( {{r_B} + {x_B}} \right)^2}\]
On plugging the values in the above relation, we get,
\[\begin{array}{c} {I_B} = \frac{{2{\rm{ kg}} \times {{\left( {0.1{\rm{ m}}} \right)}^2}}}{4} + 2{\rm{ kg}} \times {\left( {0.1{\rm{ m}} + 0.5{\rm{ m}}} \right)^2}\\ {I_B} = \frac{{{\rm{2 kg}} \times 0.01{\rm{ }}{{\rm{m}}^2}}}{4} + 2{\rm{ kg}} \times {\left( {0.6{\rm{ m}}} \right)^2}\\ {I_B} = 0.005{\rm{ kg}} \cdot {{\rm{m}}^2} + 2\;{\rm{kg}} \times {\rm{0}}{\rm{.36 }}{{\rm{m}}^2}\\ {I_B} = 0.005{\rm{ kg}} \cdot {{\rm{m}}^2} + 0.72{\rm{ kg}} \cdot {{\rm{m}}^2}\\ {I_B} = 0.725\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{array}\]
Step 5 To find the moment of inertia of slender rod, we will use parallel axis theorem,
\[I{ _r} = \frac{{{m_r} \times {L^2}}}{{12}} + {m_r}{\left( {{r_{CM}} - 0.5} \right)^2}\] ...... (1)
The cross sectional area of the given slender rod is equal at all points. Then the center of mass will be at the center of the rod.
\[\begin{array}{l} {r_{CM}} = \frac{{{l_r}}}{2}\\ {r_{CM}} = \frac{{1.5{\rm{ m}}}}{2}\\ {r_{CM}} = 0.75{\rm{ m}} \end{array}\]
Step 6 On plugging the values in equation (1), we get,
\[\begin{array}{c} {I_r} = \frac{{{\rm{4 kg}} \times {{\left( {1.5{\rm{ m}}} \right)}^2}}}{{12}} + 4{\rm{ kg}} \times {\left( {0.75{\rm{ m}} - 0.5{\rm{ m}}} \right)^2}\\ {I_r} = \frac{{{{\left( {1.5} \right)}^2}}}{3}{\rm{kg}} \cdot {{\rm{m}}^2} + 4{\rm{ kg}} \times {\left( {0.25{\rm{ m}}} \right)^2}\\ {I_r} = 0.75{\rm{ kg}} \cdot {{\rm{m}}^2} + 4{\rm{ kg}} \times 0.0625{\rm{ }}{{\rm{m}}^2}\\ {I_r} = 0.75{\rm{ kg}} \cdot {{\rm{m}}^2} + 0.25{\rm{ kg}} \cdot {{\rm{m}}^2}\\ {I_r} = 1{\rm{ kg}} \cdot {{\rm{m}}^2} \end{array}\]
Step 7 To find the mass moment of inertia for the pendulum, we will use the relation.
\[\begin{array}{c} {I_p} = {I_A} + {I_B} + {I_r}\\ {I_p} = 11.6{\rm{ kg}} \cdot {{\rm{m}}^2} + 0.725{\rm{ kg}} \cdot {{\rm{m}}^2} + 1{\rm{ kg}} \cdot {{\rm{m}}^2}\\ {I_p} = 13.325{\rm{ kg}} \cdot {{\rm{m}}^2} \end{array}\]
Step 8 To find the radius of gyration of the pendulum, we will use the relation,
\[\begin{array}{l} {I_p} = {m_p}{k^2}\\ {I_p} = \left( {{m_A} + {m_B} + {m_r}} \right){k^2} \end{array}\]
On plugging the values in the above relation, we get,
\[\begin{array}{c} 13.34{\rm{ kg}} \cdot {{\rm{m}}^2} = \left( {8{\rm{ kg}} + 2{\rm{ kg}} + 4{\rm{ kg}}} \right){k^2}\\ 13.34{\rm{ kg}} \cdot {{\rm{m}}^2} = \left( {14{\rm{ kg}}} \right){k^2}\\ {k^2} = \frac{{13.34{\rm{ kg}} \cdot {{\rm{m}}^2}}}{{{\rm{14 kg}}}}\\ k = \sqrt {\frac{{13.34}}{{14}}{\rm{ }}{{\rm{m}}^2}} \\ k = 0.976{\rm{ m}} \end{array}\]
