A body of mass m is tied to one end of a string of length l and revolves vertically in a circular path. At the lowest point of circle, what must be the K.E. Of the body so as to complete the circle

Options

A. 5 mgl
B. 4 mgl
C. `2.5 mgl
D. 2 mgl

Explanation

To solve the problem of determining the kinetic energy (K.E.) required for a body of mass \( m \) tied to a string of length \( l \) to complete a vertical circular path, we will follow these steps:

### Step 1: Understand the conditions for completing the vertical circle
To complete a vertical circle, the body must have enough kinetic energy at the highest point of the circle to maintain tension in the string. At the top of the circle, the gravitational force must provide the necessary centripetal force to keep the body moving in a circular path.

### Step 2: Analyze forces at the highest point
At the highest point of the circle, the forces acting on the body are:
- Weight (\( mg \)) acting downwards.
- Tension (\( T \)) in the string acting downwards.

The centripetal force required to keep the body in circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( r \) is the radius of the circle (which is equal to the length of the string \( l \)).

At the highest point, the equation for forces becomes:
\[ T + mg = \frac{mv^2}{l} \]
For the minimum condition to complete the circle, we set the tension \( T \) to 0:
\[ mg = \frac{mv^2}{l} \]

### Step 3: Solve for velocity at the highest point
From the equation above, we can solve for \( v^2 \):
\[ mg = \frac{mv^2}{l} \implies v^2 = gl \]
Thus, the velocity \( v \) at the highest point is:
\[ v = \sqrt{gl} \]

### Step 4: Calculate the kinetic energy at the highest point
The kinetic energy at the highest point is given by:
\[ K.E. = \frac{1}{2} mv^2 = \frac{1}{2} m(gl) = \frac{mgl}{2} \]

### Step 5: Determine the total energy at the lowest point
When the body is at the lowest point of the circle, it has both kinetic energy and potential energy. The potential energy at the lowest point is taken as zero, and the kinetic energy at this point must be enough to ensure that the body has the required energy to reach the top of the circle.

### Step 6: Apply conservation of energy
Using the conservation of mechanical energy, the total energy at the lowest point must equal the total energy at the highest point:
\[ K.E_{lowest} + PE_{lowest} = K.E_{highest} + PE_{highest} \]
At the lowest point:
- \( PE_{lowest} = 0 \)
- \( K.E_{lowest} = K.E_{initial} \)

At the highest point:
- \( PE_{highest} = mg(2l) \) (since the height is \( 2l \))
- \( K.E_{highest} = \frac{mgl}{2} \)

Thus, we have:
\[ K.E_{lowest} = \frac{mgl}{2} + mg(2l) \]
\[ K.E_{lowest} = \frac{mgl}{2} + 2mgl = \frac{mgl}{2} + \frac{4mgl}{2} = \frac{5mgl}{2} \]

### Final Answer
The kinetic energy required at the lowest point for the body to complete the vertical circle is:
\[ K.E_{lowest} = \frac{5mgl}{2} \]