A body is revolving in a vertical circle with constant mechanical energy. The speed of the body at the highest point is `sqrt(2rg)` . The speed of the body at the lowest point is

Options

A. `sqrt(7 gr)`
B. `sqrt(6gr)`
C. `sqrt(8 gr)`
D. `sqrt(9 gr)`

Answer

To solve the problem step by step, we will use the principle of conservation of mechanical energy.

### Step 1: Understand the scenario
We have a body revolving in a vertical circle. The speed at the highest point (point A) is given as \( v_h = \sqrt{2gr} \), where \( g \) is the acceleration due to gravity and \( r \) is the radius of the circle. We need to find the speed at the lowest point (point B).

### Step 2: Write down the conservation of mechanical energy
The mechanical energy at the highest point must equal the mechanical energy at the lowest point since the mechanical energy is constant. The mechanical energy consists of kinetic energy (KE) and potential energy (PE).

At the highest point (point A):
- Kinetic Energy, \( KE_h = \frac{1}{2} m v_h^2 \)
- Potential Energy, \( PE_h = mgh = mg(2r) \) (since height at the highest point is \( 2r \))

At the lowest point (point B):
- Kinetic Energy, \( KE_l = \frac{1}{2} m v_l^2 \)
- Potential Energy, \( PE_l = mgh = mg(0) \) (since height at the lowest point is \( 0 \))

### Step 3: Set up the energy conservation equation
Using the conservation of mechanical energy:
\[ KE_h + PE_h = KE_l + PE_l \]
Substituting the expressions for kinetic and potential energy:
\[ \frac{1}{2} m v_h^2 + mg(2r) = \frac{1}{2} m v_l^2 + mg(0) \]

### Step 4: Simplify the equation
We can cancel \( m \) from all terms (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_h^2 + 2gr = \frac{1}{2} v_l^2 \]

### Step 5: Substitute \( v_h \) and solve for \( v_l \)
We know \( v_h = \sqrt{2gr} \), so we substitute this into the equation:
\[ \frac{1}{2} (\sqrt{2gr})^2 + 2gr = \frac{1}{2} v_l^2 \]
Calculating \( (\sqrt{2gr})^2 \):
\[ \frac{1}{2} (2gr) + 2gr = \frac{1}{2} v_l^2 \]
This simplifies to:
\[ gr + 2gr = \frac{1}{2} v_l^2 \]
\[ 3gr = \frac{1}{2} v_l^2 \]

### Step 6: Solve for \( v_l^2 \)
Multiplying both sides by 2:
\[ 6gr = v_l^2 \]

### Step 7: Find \( v_l \)
Taking the square root of both sides gives:
\[ v_l = \sqrt{6gr} \]

### Final Answer
The speed of the body at the lowest point is \( v_l = \sqrt{6gr} \).

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