A particle rests on the top of a hemishpere of radius R. Find the smallest horizontal velocity that must be imparted to the particle if it is yo leave the hemisphere without sliding down it.

Options

A. `sqrt(gR )`
B. `sqrt(2 gR)`
C. `sqrt(3gR)`
D. `sqrt(5gR)`

Explanation

To solve the problem of finding the smallest horizontal velocity that must be imparted to a particle resting on the top of a hemisphere of radius \( R \) so that it leaves the hemisphere without sliding down, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces**:
When the particle is at the top of the hemisphere, it experiences two main forces:
- The gravitational force acting downward, which is \( mg \) (where \( m \) is the mass of the particle and \( g \) is the acceleration due to gravity).
- The centrifugal force due to the horizontal velocity \( v \) imparted to the particle.

2. **Centrifugal Force Calculation**:
The centrifugal force acting on the particle when it moves in a circular path can be expressed as:
\[ F_{cf} = \frac{mv^2}{R} \]
where \( R \) is the radius of the hemisphere.

3. **Condition for Leaving the Hemisphere**:
For the particle to leave the hemisphere without sliding down, the centrifugal force must be equal to the gravitational force acting on the particle at the point it leaves the hemisphere. Therefore, we set:
\[ F_{cf} = mg \]
This gives us the equation:
\[ \frac{mv^2}{R} = mg \]

4. **Simplifying the Equation**:
We can cancel the mass \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{v^2}{R} = g \]

5. **Solving for Velocity**:
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = gR \]
Taking the square root of both sides gives us the velocity:
\[ v = \sqrt{gR} \]

6. **Conclusion**:
Thus, the smallest horizontal velocity that must be imparted to the particle for it to leave the hemisphere without sliding down is:
\[ v = \sqrt{gR} \]

### Final Answer:
The correct answer is \( \sqrt{gR} \).

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