A particle is moving in a circular path of radius r under the action of a force F. If at an instant velocity of particle is v, and speed of particle is increasing, then

Options

A. `vecF.vecv=0`
B. `vecF.vecv gt 0`
C. `vecF.vecv lt0`
D. `vecF.vecvge0`

Answer

To solve the problem, we need to analyze the motion of a particle moving in a circular path under the influence of a force \( \mathbf{F} \). Given that the speed of the particle is increasing, we can derive the relationship between the force vector \( \mathbf{F} \) and the velocity vector \( \mathbf{v} \).

### Step-by-step Solution:

1. **Understanding the Motion**:
The particle is moving in a circular path of radius \( r \). The velocity \( \mathbf{v} \) is tangential to the circular path, and there is a centripetal acceleration \( \mathbf{a_c} \) directed towards the center of the circle.

2. **Components of Acceleration**:
Since the speed of the particle is increasing, there is also a tangential acceleration \( \mathbf{a_t} \) in the direction of the velocity vector \( \mathbf{v} \). The total acceleration \( \mathbf{a} \) can be expressed as:
\[ \mathbf{a} = \mathbf{a_t} + \mathbf{a_c} \]
where \( \mathbf{a_c} \) is directed towards the center of the circular path.

3. **Direction of Force**:
According to Newton's second law, the net force \( \mathbf{F} \) acting on the particle is equal to the mass \( m \) of the particle multiplied by its total acceleration:
\[ \mathbf{F} = m \mathbf{a} \]
The force \( \mathbf{F} \) will have components in both the tangential direction (due to \( \mathbf{a_t} \)) and the radial direction (due to \( \mathbf{a_c} \)).

4. **Dot Product of Force and Velocity**:
We need to evaluate the dot product \( \mathbf{F} \cdot \mathbf{v} \). The dot product can be expressed as:
\[ \mathbf{F} \cdot \mathbf{v} = |\mathbf{F}| |\mathbf{v}| \cos \theta \]
where \( \theta \) is the angle between the force vector \( \mathbf{F} \) and the velocity vector \( \mathbf{v} \).

5. **Analyzing the Angle**:
Since the speed is increasing, the tangential component of the force \( \mathbf{F} \) must be in the same direction as the velocity \( \mathbf{v} \). This means that the angle \( \theta \) between \( \mathbf{F} \) and \( \mathbf{v} \) is between \( 0^\circ \) and \( 90^\circ \). Therefore, \( \cos \theta \) is positive.

6. **Conclusion**:
Since both \( |\mathbf{F}| \) and \( |\mathbf{v}| \) are positive and \( \cos \theta > 0 \), we conclude that:
\[ \mathbf{F} \cdot \mathbf{v} > 0 \]
Thus, the correct answer is **Option 2**: \( \mathbf{F} \cdot \mathbf{v} > 0 \).