Let's calculate each part of the problem step by step:

Given:

• Radius of sphere $R = 0.64 \, \text{m}$
• Mass of sphere $m_{\text{sphere}} = 0.40 \, \text{kg}$
• Length of rod $L = 1.20 \, \text{m}$
• Mass of rod $m_{\text{rod}} = 0.52 \, \text{kg}$
• Pivot distance $X = 0.42 \, \text{m}$
• Acceleration due to gravity $g = 9.81 \, \text{m/s}^2$

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(a) Moment of Inertia

1. For the rod: $I_{\text{rod}} = \frac{1}{12} m_{\text{rod}} L^2 + m_{\text{rod}} X^2$ where $\frac{1}{12} m_{\text{rod}} L^2$ is the moment of inertia about its center, and the second term accounts for the distance $X$ from the pivot (parallel axis theorem).

2. For the hollow sphere: The moment of inertia of the hollow sphere about its center is: $I_{\text{sphere-center}} = \frac{2}{3} m_{\text{sphere}} R^2$ Using the parallel axis theorem to shift the axis to the pivot (which is at a distance $(L - X)$ from the center of the sphere): $I_{\text{sphere}} = I_{\text{sphere-center}} + m_{\text{sphere}} (L - X)^2$

   Adding the contributions from both the rod and the sphere: $I_{\text{total}} = I_{\text{rod}} + I_{\text{sphere}}$ Result: $I_{\text{total}} = 0.507 \, \text{kg} \cdot \text{m}^2$

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(b) Torque about the Fulcrum

Torque is calculated as $\tau = r \cdot F \sin \theta$. In this case, the force is the weight due to gravity, and since the forces are vertical, $\sin 90^\circ = 1$.

1. Torque due to the rod: $\tau_{\text{rod}} = m_{\text{rod}} g (X - \text{center of mass distance})$ The center of mass of the rod is at $\frac{L}{2}$, so: $\tau_{\text{rod}} = m_{\text{rod}} g (X - \frac{L}{2})$

2. Torque due to the sphere: The force due to the sphere acts at distance $(L - X)$ from the pivot: $\tau_{\text{sphere}} = m_{\text{sphere}} g (L - X)$

Summing both torques: $\tau_{\text{total}} = \tau_{\text{rod}} + \tau_{\text{sphere}}$ Result: $\tau_{\text{total}} = 2.14 \, \text{N} \cdot \text{m}$

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(c) Angular Acceleration

Using Newton's second law for rotation, the angular acceleration $\alpha$ is: $\alpha = \frac{\tau_{\text{net}}}{I_{\text{total}}}$ Substituting the values for torque and moment of inertia: $\alpha = \frac{2.14}{0.507} = 4.23 \, \text{rad/s}^2$

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(d) Linear Acceleration of the Right End of the Rod

The linear acceleration $a$ of the right end of the rod is related to the angular acceleration by: $a = \alpha \times L$ Substituting the angular acceleration and length of the rod: $a = 4.23 \times 1.20 = 5.07 \, \text{m/s}^2$

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Final Results:

• (a) Moment of inertia: $I_{\text{total}} = 0.507 \, \text{kg} \cdot \text{m}^2$
• **(b)