A particle is rotated in vertical circle by connecting it to string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is

Options

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

Answer

To find the minimum speed of a particle at the horizontal position for it to complete a vertical circle, we can follow these steps:

### Step 1: Understand the Forces at the Top of the Circle
When the particle reaches the top of the vertical circle, the forces acting on it are:
- The gravitational force (mg) acting downwards.
- The tension (T) in the string acting downwards.

For the particle to just complete the circle, the tension in the string can be zero at the topmost point. Thus, the centripetal force required to keep the particle in circular motion is provided entirely by the weight of the particle.

### Step 2: Apply the Centripetal Force Condition
At the top of the circle, the centripetal force is given by:
\[ \frac{mv^2}{r} = mg \]
Here, \(v\) is the speed of the particle at the top of the circle, \(m\) is the mass of the particle, and \(r\) is the radius of the circle.

From this equation, we can derive:
\[ v^2 = gr \]
This means that the minimum speed \(v\) at the top of the circle must be:
\[ v = \sqrt{gr} \]

### Step 3: Energy Conservation from the Bottom to the Top
Now, we need to consider the energy conservation from the horizontal position (where the particle is released) to the top of the circle. The potential energy at the top is higher than at the horizontal position.

The height \(h\) gained when moving from the horizontal position to the top of the circle is equal to the diameter of the circle, which is \(2r\).

The work done against gravity when moving to the top is:
\[ \Delta PE = mg(2r) \]

The change in kinetic energy as it moves from the horizontal position to the top is given by:
\[ \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \]
where \(u\) is the initial speed at the horizontal position.

### Step 4: Set Up the Energy Conservation Equation
Using the principle of conservation of energy, we can write:
\[ \Delta KE + \Delta PE = 0 \]
This leads to:
\[ \frac{1}{2}mv^2 - \frac{1}{2}mu^2 - mg(2r) = 0 \]

### Step 5: Substitute for \(v\) and Rearrange
Substituting \(v^2 = gr\) into the energy equation gives:
\[ \frac{1}{2}m(gr) - \frac{1}{2}mu^2 - mg(2r) = 0 \]
This simplifies to:
\[ \frac{1}{2}mgr - \frac{1}{2}mu^2 - 2mgr = 0 \]
Combining terms leads to:
\[ -\frac{3}{2}mgr = \frac{1}{2}mu^2 \]

### Step 6: Solve for \(u\)
Rearranging gives:
\[ u^2 = 3gr \]
Taking the square root gives:
\[ u = \sqrt{3gr} \]

### Final Answer
Thus, the minimum speed \(u\) of the particle when the string is horizontal for it to complete the circle is:
\[ u = \sqrt{3gr} \]

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