A hollow sphere is attached to the end of a uniform rod. The sphere has a radius of 0.64 m and a mass of 0.40 kg. The rod has a length of 1.20 m and a mass of 0.52 kg. The rod is placed on a fulcrum (pivot) at *X* = 0.42 m from the left end of the rod.

(a) Calculate the moment of inertia (click for graphical table) of the contraption around the fulcrum.
  kg m2

(b) Calculate the torque about the fulcrum, using CCW as positive.
  N m

(c) Calculate the angular acceleration of the contraption, using CCW as positive.
  rad/s2

(d) Calculate the linear acceleration of the right end of the rod, using up as positive.
  m/s2

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** \( g = 9.81 \, \text{m/s}^2 \)

---

### (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:**
   \[
   I_{\text{total}} = 0.507 \, \text{kg} \cdot \text{m}^2
   \]

---

### (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:**
\[
\tau_{\text{total}} = 2.14 \, \text{N} \cdot \text{m}
\]

---

### (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:
\[
\alpha = \frac{2.14}{0.507} = 4.23 \, \text{rad/s}^2
\]

---

### (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:
\[
a = 4.23 \times 1.20 = 5.07 \, \text{m/s}^2
\]

---

### Final Results:
- **(a) Moment of inertia**: \( I_{\text{total}} = 0.507 \, \text{kg} \cdot \text{m}^2 \)
- **(b)

A hollow sphere is attached to the end of a uniform rod. The sphere has a radius of 0.64 m and a mass of 0.40 kg. The rod has a length of 1.20 m and a mass of 0.52 kg. The rod is placed on a fulcrum (pivot) at X = 0.42 m from the left end of the rod. (a) Calculate the moment of inertia of the contraption around the fulcrum. (b) Calculate the torque about the fulcrum, using CCW as positive. (c) Calculate the angular acceleration of the contraption, using CCW as positive. (d) Calculate the linear acceleration of the right end of the rod, using up as positive.

Solve a physics problem involving a hollow sphere and rod system. Calculate the moment of inertia, torque, angular acceleration, and linear acceleration using key formulas. This problem is suitable for undergraduate physics students and covers rotational dynamics.

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