A smooth sphere of radius R is made to translate line with a constant acceleration a=g. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of angle θ as it slides down.

Explanation

To solve the problem, we need to determine the speed of a particle sliding down a smooth sphere that is translating with a constant acceleration equal to \( g \). The particle is released from the top of the sphere at zero velocity with respect to the sphere. We will use the work-energy theorem to find the speed of the particle as a function of the angle \( \theta \).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Particle**:
- The gravitational force \( mg \) acts downward.
- A pseudo force \( F_{pseudo} = ma \) acts horizontally in the opposite direction of the sphere's acceleration. Since \( a = g \), this force becomes \( mg \).

2. **Set Up the Coordinate System**:
- Let the center of the sphere be at the origin. When the particle is at an angle \( \theta \) from the vertical, its position can be described in terms of the radius \( R \) of the sphere:
- Horizontal position: \( x = R \sin \theta \)
- Vertical position: \( y = R (1 - \cos \theta) \)

3. **Calculate the Work Done**:
- The work done by gravity as the particle moves down is:
\[ W_{gravity} = mg \cdot h = mg \cdot (R(1 - \cos \theta)) \]
- The work done by the pseudo force as the particle moves horizontally is:
\[ W_{pseudo} = F_{pseudo} \cdot d = mg \cdot (R \sin \theta) \]

4. **Apply the Work-Energy Theorem**:
- According to the work-energy theorem:
\[ W_{gravity} + W_{pseudo} = \Delta KE \]
- The change in kinetic energy is given by:
\[ \Delta KE = \frac{1}{2} m V^2 - 0 = \frac{1}{2} m V^2 \]
- Therefore, we can write:
\[ mg(R(1 - \cos \theta)) + mg(R \sin \theta) = \frac{1}{2} m V^2 \]

5. **Simplify the Equation**:
- Cancel \( m \) from both sides:
\[ g(R(1 - \cos \theta) + g(R \sin \theta)) = \frac{1}{2} V^2 \]
- This simplifies to:
\[ gR(1 - \cos \theta + \sin \theta) = \frac{1}{2} V^2 \]

6. **Solve for \( V^2 \)**:
- Rearranging gives:
\[ V^2 = 2gR(1 - \cos \theta + \sin \theta) \]

7. **Find \( V \)**:
- Taking the square root:
\[ V = \sqrt{2gR(1 - \cos \theta + \sin \theta)} \]

### Final Answer:
The speed of the particle with respect to the sphere as a function of angle \( \theta \) is:
\[ V = \sqrt{2gR(1 - \cos \theta + \sin \theta)} \]