A bead can slide on a smooth circular wire frame of radius R which is fixed in a vartical plane. The bead is displaced slightly 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 vertical line is

Options

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

Answer

To find the speed of the bead as a function of the angle θ made with the vertical line, we can use the principle of conservation of mechanical energy. Here are the steps to derive the speed:

### Step 1: Understand the Initial and Final States
- Initially, the bead is at the highest point of the circular wire frame (point A) and is at rest. Therefore, its initial kinetic energy (KE) is 0, and its potential energy (PE) is maximum.
- When the bead is displaced to an angle θ (point B), it has some speed (v) and a lower potential energy.

### Step 2: Define the Height Change
- The height (h) of the bead when it is at angle θ can be calculated as:
\[ h = R - R \cos \theta = R(1 - \cos \theta) \]
Here, R is the radius of the circular wire frame.

### Step 3: Write the Conservation of Energy Equation
- According to the conservation of energy:
\[ \text{Total Energy at A} = \text{Total Energy at B} \]
\[ KE_A + PE_A = KE_B + PE_B \]
- At point A (highest point):
- \( KE_A = 0 \)
- \( PE_A = mgR \) (taking the reference point at B)

- At point B (displaced point):
- \( KE_B = \frac{1}{2} mv^2 \)
- \( PE_B = 0 \) (taking the potential energy at point B as zero)

### Step 4: Set Up the Equation
- Plugging in the values:
\[ 0 + mgR = \frac{1}{2} mv^2 + 0 \]
- This simplifies to:
\[ mgR = \frac{1}{2} mv^2 \]

### Step 5: Cancel Mass and Rearrange
- Cancel the mass (m) from both sides (assuming m ≠ 0):
\[ gR = \frac{1}{2} v^2 \]
- Rearranging gives:
\[ v^2 = 2gR \]

### Step 6: Substitute for Height
- Now, substituting the height (h) we found earlier:
\[ v^2 = 2g(R(1 - \cos \theta)) \]
- Therefore, we have:
\[ v = \sqrt{2gR(1 - \cos \theta)} \]

### Final Result
The speed of the bead as a function of the angle θ is:
\[ v = \sqrt{2gR(1 - \cos \theta)} \]