A bead is arranged to move with constant speed around a loop that lies in a vetical plane. The magnitude of the net force on the bead is

Options

A. maximum at the bottom
B. maximum at the top
C. maximum at the side points
D. the same at all points

Answer

To determine the magnitude of the net force on a bead moving with constant speed around a vertical loop, we can analyze the forces acting on the bead at different points in the loop.

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Bead**:
- The bead experiences two main forces: the gravitational force (weight) acting downward, \( mg \), where \( m \) is the mass of the bead and \( g \) is the acceleration due to gravity, and the normal force \( N \) exerted by the track on the bead.

2. **Analyze the Bead at Different Points**:
- We will analyze the forces at three key points: the bottom of the loop, the top of the loop, and a point on the side (horizontal position).

3. **At the Bottom of the Loop**:
- At this position, the net force towards the center (centripetal force) is given by:
\[ N_1 - mg = \frac{mv^2}{R} \]
Rearranging gives:
\[ N_1 = mg + \frac{mv^2}{R} \]
- Here, \( N_1 \) is the normal force at the bottom, which is the maximum force experienced by the bead.

4. **At the Top of the Loop**:
- At the top, the forces acting on the bead are:
\[ N_2 + mg = \frac{mv^2}{R} \]
Rearranging gives:
\[ N_2 = \frac{mv^2}{R} - mg \]
- Here, \( N_2 \) is the normal force at the top, which is the minimum force experienced by the bead.

5. **At the Horizontal Position**:
- At this position, the forces are:
\[ N_3 = \frac{mv^2}{R} \]
- Here, \( N_3 \) is the normal force at the horizontal position.

6. **Comparison of Forces**:
- From the equations derived:
- At the bottom: \( N_1 = mg + \frac{mv^2}{R} \) (maximum)
- At the top: \( N_2 = \frac{mv^2}{R} - mg \) (minimum)
- At the horizontal position: \( N_3 = \frac{mv^2}{R} \)

7. **Conclusion**:
- The magnitude of the net force on the bead is maximum at the bottom of the loop, where it is given by:
\[ N_1 = mg + \frac{mv^2}{R} \]

### Final Answer:
The magnitude of the net force on the bead is maximum at the bottom of the loop, given by \( N_1 = mg + \frac{mv^2}{R} \).