What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

Options

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

Answer

To find the minimum velocity \( u \) with which a body of mass \( m \) must enter a vertical loop of radius \( R \) to complete the loop, we can follow these steps:

### Step 1: Analyze the forces at the top of the loop
At the highest point of the loop, the forces acting on the body are:
- The gravitational force \( mg \) acting downward.
- The centripetal force required to keep the body moving in a circular path, which is provided by the gravitational force and the tension in the string (if any). For the minimum condition, we can assume the tension is zero.

At the top of the loop, the centripetal force is given by:
\[ \frac{mv^2}{R} \]
where \( v \) is the velocity at the top of the loop.

For the body to just complete the loop, the gravitational force must provide the necessary centripetal force:
\[ mg = \frac{mv^2}{R} \]
This simplifies to:
\[ g = \frac{v^2}{R} \]
From this, we can express \( v \) as:
\[ v = \sqrt{gR} \]

### Step 2: Apply conservation of mechanical energy
Now, we will use the conservation of mechanical energy to find the minimum velocity \( u \) at the bottom of the loop. The total mechanical energy at the bottom must equal the total mechanical energy at the top.

At the bottom of the loop:
- Potential energy \( U_A = 0 \) (taking this point as the reference level).
- Kinetic energy \( K_A = \frac{1}{2}mu^2 \).

At the top of the loop:
- Potential energy \( U_B = mg(2R) \) (the height is \( 2R \)).
- Kinetic energy \( K_B = \frac{1}{2}mv^2 \).

Using conservation of energy:
\[ U_A + K_A = U_B + K_B \]
Substituting the values:
\[ 0 + \frac{1}{2}mu^2 = mg(2R) + \frac{1}{2}mv^2 \]

### Step 3: Substitute \( v \) from Step 1
Now, substituting \( v = \sqrt{gR} \) into the energy equation:
\[ \frac{1}{2}mu^2 = mg(2R) + \frac{1}{2}m(\sqrt{gR})^2 \]
This simplifies to:
\[ \frac{1}{2}mu^2 = mg(2R) + \frac{1}{2}mgR \]
\[ \frac{1}{2}mu^2 = mg(2R + \frac{1}{2}R) \]
\[ \frac{1}{2}mu^2 = mg(\frac{5}{2}R) \]

### Step 4: Solve for \( u \)
Now, we can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2}u^2 = g\frac{5}{2}R \]
Multiplying both sides by 2:
\[ u^2 = 5gR \]
Taking the square root:
\[ u = \sqrt{5gR} \]

### Final Answer
The minimum velocity \( u \) with which a body of mass \( m \) must enter a vertical loop of radius \( R \) to complete the loop is:
\[ \boxed{u = \sqrt{5gR}} \]