A bead can slide on a smooth circular wire frame of radius r which is fixed in a vertical plane. The bead is displaced slighty from the highest point of the wire frame. The speed of the bead subsequently as a function of the angle `theta` made by the bead with the verticle line is

Options

A. `sqrt(2gr)`
B. `sqrt(2gr(1-sintheta))`
C. `sqrt(2gr(1-costheta))`
D. `2sqrt(gr)`

Explanation

To solve the problem of finding the speed of the bead as a function of the angle θ made with the vertical line, we can follow these steps:

### Step 1: Understand the Setup
The bead is on a smooth circular wire of radius \( r \) and is displaced slightly from the highest point. As it moves down, it loses potential energy and gains kinetic energy.

### Step 2: Determine the Height \( h \)
When the bead is at an angle \( \theta \) with the vertical, the height \( h \) of the bead above the lowest point of the circular path can be expressed as:
\[ h = r(1 - \cos \theta) \]
This equation comes from the geometry of the circle, where the vertical height from the top to the bead's position is given by the difference in the radius and the vertical component of the radius at angle \( \theta \).

### Step 3: Apply Conservation of Energy
According to the principle of conservation of energy, the decrease in potential energy equals the increase in kinetic energy. The potential energy lost when the bead moves from the top to the height \( h \) is:
\[ \Delta PE = mg h = mg \cdot r(1 - \cos \theta) \]
The kinetic energy gained by the bead is:
\[ KE = \frac{1}{2} mv^2 \]
Setting the decrease in potential energy equal to the increase in kinetic energy gives us:
\[ mgr(1 - \cos \theta) = \frac{1}{2} mv^2 \]

### Step 4: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gr(1 - \cos \theta) = \frac{1}{2} v^2 \]

### Step 5: Solve for \( v \)
To find the speed \( v \), we can rearrange the equation:
\[ v^2 = 2gr(1 - \cos \theta) \]
Taking the square root of both sides gives us:
\[ v = \sqrt{2gr(1 - \cos \theta)} \]

### Conclusion
Thus, the speed of the bead as a function of the angle \( \theta \) is:
\[ v = \sqrt{2gr(1 - \cos \theta)} \]