Two thin rods of negligible mass are rigidly attached at their ends to form a \(90^{\circ}\) angle. The rods rotate in the \(x y\) plane with the joined ends forming the pivot at the origin. A particle of mass \(75 \mathrm{~g}\) is attached to one rod a distance of \(42 \mathrm{~cm}\) from the origin, and a particle of mass \(30 \mathrm{~g}\) is attached to the other rod a distance of \(65 \mathrm{~cm}\) from the origin. ( \(a\) ) What is the rotational inertia of the assembly? \((b)\) How would the rotational inertia change if the particles were both attached to one rod at the given distances from the origin?

Using the formula for calculating rotational inertia, which is \(I = m*r^{2}\) for each particle, where \(m\) is mass, and \(r\) is distance from the pivot point. The total rotational inertia of the assembly is the sum of the moment of inertias of both particles. For the particle with mass \(m_1 = 75 \, g = 0.075 \, kg\) at distance \(r_1 = 42 \, cm = 0.42 \, m\), the rotational inertia \(I_1 = m_1 * r_1^{2} = 0.075 \, kg * (0.42 \, m)^{2}\). Solving this yields \(I_1 \approx 0.01305 \, kg \cdot m^{2}\). The particle with mass \(m_2 = 30 \, g = 0.030 \, kg\) at distance \(r_2 = 65 \, cm = 0.65 \, m\), has rotational inertia \(I_2 = m_2 * r_2^{2} = 0.030 \, kg * (0.65 \, m)^{2}\). Solving this yields \(I_2 \approx 0.012825\, kg \cdot m^{2}\). Therefore, the total rotational inertia of the system, \(I = I_1 + I_2 \approx 0.01305 \, kg \cdot m^{2} + 0.012825 \, kg \cdot m^{2} \approx 0.025875 \, kg \cdot m^{2}\).

In this configuration, both the particles are located on the same rod. Therefore, we can simply apply the formula for each particle separately and add them together as done in step 1. When we calculate, we find \(I' = I_1 + I_2 \approx 0.025875 \, kg \cdot m^{2}\), which is the same value as calculated in the first configuration.