91Ó°ÊÓ

First, let's calculate the initial potential energy of the pendulum when it is released from rest at an angle of \(30^{\circ}\). The distance the center of mass of the rod and sphere rise when the pendulum is released is given by the sum of the vertical displacements of the center of mass of the rod and the sphere. Let's denote the height change of the rod as \(h_{rod}\) and the height change of the sphere as \(h_{sphere}\). For the rod: \(h_{rod} = \frac{1}{2} \cdot 1 \cdot (1 - \cos(30^{\circ})) = \frac{1}{4}(1 - \frac{\sqrt{3}}{2})\) m For the sphere: \(h_{sphere} = 1 - 1 \cdot \cos(30^{\circ}) = 1 - \frac{\sqrt{3}}{2}\) m Now, calculate the potential energy stored in the system: \(U = m_{rod} gh_{rod} + m_{sphere} gh_{sphere}\) \(U = (2\text{kg})(9.81\frac{\text{m}}{\text{s}^2})\cdot\frac{1}{4}(1 - \frac{\sqrt{3}}{2})\text{m} + (0.3\text{kg})(9.81\frac{\text{m}}{\text{s}^2})(1-\frac{\sqrt{3}}{2})\text{m}\) \(U \approx 3.7844\text{J}\)

Next, we need to find the moment of inertia of the system, which consists of the rod and the sphere. The total moment of inertia is the sum of the moment of inertia of the rod and the sphere. For a rod pivoted at one end, the moment of inertia is given by: \(I_{rod} = \frac{1}{3}m_{rod}L^2\) For a solid sphere pivoted at a distance L from its center, the moment of inertia can be calculated with the parallel axis theorem: \(I_{sphere} = \frac{2}{5}m_{sphere}R^2 + m_{sphere}L^2\) The total moment of inertia is: \(I = I_{rod} + I_{sphere}\)

Now we apply the conservation of energy principle. The initial potential energy will be converted into the rotational kinetic energy at the lowest point: \(U = \frac{1}{2}I\omega^2\) Where \(\omega\) is the angular velocity at the lowest point. We can rearrange to isolate \(\omega\): \(\omega^2 = \frac{2U}{I}\)

Finally, we plug in the values for \(U\) and \(I\) to find the angular velocity: \(\omega^2 = \frac{2(3.7844\text{J})}{\frac{1}{3}(2\text{kg})(1\text{m})^2 + \frac{2}{5}(0.3\text{kg})(0.2\text{m})^2 + (0.3\text{kg})(1\text{m})^2}\) \(\omega \approx 3.052\frac{\text{rad}}{\text{s}}\) Thus, the angular velocity of the pendulum at the lowest point is approximately \(3.052\frac{\text{rad}}{\text{s}}\).