A body is mass `m` is rotating in a vertical circle of radius 'r' with critical speed. The difference in its `K.E` at the top and at the bottom is

Options

A. mgr
B. 2mgr
C. `(mg)/(r )`
D. `(2mg)/(r )`

Answer

To find the difference in kinetic energy (K.E) of a body of mass \( m \) rotating in a vertical circle of radius \( r \) at critical speed, we can follow these steps:

### Step 1: Understand the positions in the vertical circle
- The body is at two positions: the top of the circle and the bottom of the circle.
- Let’s denote the velocity at the top as \( v_1 \) and at the bottom as \( v_2 \).

### Step 2: Calculate the height difference
- The height difference between the top and the bottom of the circle is \( 2r \) (since the radius is \( r \) at the top and \( r \) at the bottom).

### Step 3: Apply the Work-Energy Theorem
- According to the work-energy theorem, the work done by gravity is equal to the change in kinetic energy as the body moves from the top to the bottom.
- The work done by gravity when moving from the top to the bottom is given by:
\[ W = -mg \cdot (2r) \]
Here, the negative sign indicates that the work done by gravity is in the opposite direction to the displacement.

### Step 4: Relate work done to change in kinetic energy
- The change in kinetic energy (\( \Delta KE \)) can be expressed as:
\[ W = KE_{bottom} - KE_{top} \]
Therefore, we have:
\[ -mg \cdot (2r) = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 \]

### Step 5: Rearranging the equation
- Rearranging the equation gives us:
\[ \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 = -2mgr \]
- Taking the absolute value of the change in kinetic energy, we find:
\[ \Delta KE = \left| KE_{bottom} - KE_{top} \right| = 2mgr \]

### Final Result
- The difference in kinetic energy at the top and at the bottom is:
\[ \Delta KE = 2mgr \]