A particle of mass m is kept on a fixed, smooth sphere of radius R at a position, where the radius through the particle makes an angle of 30 ∘ with the vertical. The particle is released from this position. (a) What is the force exerted by the sphere on the particle just after the release? (b) Find the distance traveled by the particle before it leaves contact with the sphere.

Explanation

To solve the problem, we will break it down into two parts as specified in the question.

### Part (a): Force exerted by the sphere on the particle just after release

1. **Identify the forces acting on the particle**:
- The gravitational force \( mg \) acts downward.
- The normal force \( N \) exerted by the sphere acts perpendicular to the surface of the sphere.

2. **Determine the angle**:
- The particle is at an angle of \( 30^\circ \) with the vertical. Therefore, the angle with the horizontal is \( 90^\circ - 30^\circ = 60^\circ \).

3. **Resolve the gravitational force**:
- The component of the gravitational force acting along the radius (towards the center of the sphere) is \( mg \cos(30^\circ) \).
- The component of the gravitational force acting perpendicular to the radius (along the surface) is \( mg \sin(30^\circ) \).

4. **Set up the equation for forces**:
- Since the particle is momentarily at rest just after release, the net force in the radial direction must equal the centripetal force required for circular motion. However, since the velocity is zero, the centripetal force is also zero.
- Therefore, we can write:
\[ N = mg \cos(30^\circ) \]

5. **Calculate \( N \)**:
- Using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ N = mg \cdot \frac{\sqrt{3}}{2} \]

### Part (b): Distance traveled by the particle before it leaves contact with the sphere

1. **Identify the point of leaving contact**:
- The particle will leave contact when the normal force \( N \) becomes zero.

2. **Set up the equation for forces at the point of leaving contact**:
- At the point of leaving contact, the only force providing the centripetal acceleration is the component of the weight acting towards the center of the sphere:
\[ mg \cos(\theta) = \frac{mv^2}{R} \]
- Here, \( \theta \) is the angle at which the particle leaves contact and \( v \) is the speed of the particle at that point.

3. **Use conservation of energy**:
- The potential energy lost by the particle as it moves from the initial height to the height at angle \( \theta \) is converted into kinetic energy:
\[ mgR \cos(30^\circ) - mgR \cos(\theta) = \frac{1}{2} mv^2 \]

4. **Solve for \( v^2 \)**:
- Rearranging gives:
\[ mgR \left( \cos(30^\circ) - \cos(\theta) \right) = \frac{1}{2} mv^2 \]
- Canceling \( m \) and multiplying by 2 gives:
\[ 2gR \left( \cos(30^\circ) - \cos(\theta) \right) = v^2 \]

5. **Substituting \( v^2 \) into the force equation**:
- Substitute \( v^2 \) into the centripetal force equation:
\[ g \cos(\theta) = \frac{2gR \left( \cos(30^\circ) - \cos(\theta) \right)}{R} \]
- Simplifying gives:
\[ \cos(\theta) = 2 \left( \cos(30^\circ) - \cos(\theta) \right) \]

6. **Solve for \( \cos(\theta) \)**:
- Rearranging gives:
\[ 3 \cos(\theta) = 2 \cos(30^\circ) \]
- Thus,
\[ \cos(\theta) = \frac{2 \cdot \frac{\sqrt{3}}{2}}{3} = \frac{\sqrt{3}}{3} \]

7. **Find the angle \( \theta \)**:
- Therefore,
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

8. **Calculate the distance traveled**:
- The distance traveled along the arc from the initial position to the point of leaving contact is given by:
\[ s = R \left( \theta - \frac{\pi}{6} \right) \]

### Final Answers:
(a) The force exerted by the sphere on the particle just after release is:
\[ N = mg \cdot \frac{\sqrt{3}}{2} \]

(b) The distance traveled by the particle before it leaves contact with the sphere is:
\[ s = R \left( \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) - \frac{\pi}{6} \right) \]