A smooth sphere of radius R is made to translate oin a straight line with a constant acceleration a. A particle kept on the top of the sphere is released rom there at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle `theta` it slides.

Explanation

To solve the problem, we need to analyze the motion of a particle sliding off the top of a sphere that is translating with a constant acceleration. We will derive the speed of the particle with respect to the sphere as it slides down, as a function of the angle \( \theta \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a smooth sphere of radius \( R \) that is translating with a constant acceleration \( a \).
- A particle is placed at the top of the sphere and is released from rest with respect to the sphere.

2. **Defining the Motion**:
- As the particle slides down, it moves along the surface of the sphere. We need to find the speed of the particle with respect to the sphere at an angle \( \theta \) from the vertical.

3. **Displacement Analysis**:
- The vertical displacement \( x \) of the particle from the top of the sphere to the position at angle \( \theta \) can be expressed as:
\[ x = R - R \cos \theta = R(1 - \cos \theta) \]
- The horizontal displacement \( d \) can be expressed as:
\[ d = R \sin \theta \]

4. **Forces Acting on the Particle**:
- The forces acting on the particle include:
- The gravitational force \( mg \) acting downward.
- A pseudo force \( ma \) acting horizontally backward due to the acceleration of the sphere.

5. **Calculating Work Done**:
- The work done by the pseudo force as the particle moves horizontally is:
\[ W_{\text{pseudo}} = (ma)(R \sin \theta) \]
- The work done by the gravitational force as the particle moves vertically is:
\[ W_{\text{gravity}} = mg(R(1 - \cos \theta)) \]

6. **Applying Work-Energy Theorem**:
- According to the work-energy theorem, the total work done is equal to the change in kinetic energy:
\[ W_{\text{total}} = W_{\text{pseudo}} + W_{\text{gravity}} = \Delta KE \]
- The change in kinetic energy is given by:
\[ \Delta KE = \frac{1}{2} mv^2 - 0 = \frac{1}{2} mv^2 \]

7. **Setting Up the Equation**:
- Combining the work done:
\[ ma(R \sin \theta) + mg(R(1 - \cos \theta)) = \frac{1}{2} mv^2 \]
- Dividing through by \( m \):
\[ a(R \sin \theta) + g(R(1 - \cos \theta)) = \frac{1}{2} v^2 \]

8. **Solving for \( v \)**:
- Rearranging gives:
\[ v^2 = 2aR \sin \theta + 2gR(1 - \cos \theta) \]
- Taking the square root:
\[ v = \sqrt{2aR \sin \theta + 2gR(1 - \cos \theta)} \]

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