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

Explanation

To find 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, we can follow these steps:

### Step 1: Understand the Problem
We need to determine the minimum velocity at the bottom of the loop such that when the body reaches the top of the loop, it has just enough speed to maintain contact with the loop.

### Step 2: Analyze Forces at the Top of the Loop
At the top of the loop, the forces acting on the body are:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting downward (which will be zero at the minimum velocity).

For the body to just maintain contact at the top of the loop, the centripetal force required to keep the body in circular motion must equal the gravitational force acting on it. Thus, we can write:
\[ mg + N = \frac{mv^2}{R} \]
Setting \( N = 0 \) (for the minimum velocity), we have:
\[ mg = \frac{mv^2}{R} \]
This simplifies to:
\[ v^2 = gR \]

### Step 3: Apply Conservation of Energy
We will apply the work-energy theorem from the bottom of the loop to the top of the loop. The change in potential energy as the body moves from the bottom to the top is:
\[ \Delta PE = mg(2R) = 2mgR \]
The change in kinetic energy is given by:
\[ \Delta KE = \frac{1}{2}mv_{\text{top}}^2 - \frac{1}{2}mv_{\text{min}}^2 \]

### Step 4: Set Up the Energy Equation
Using the conservation of energy, we have:
\[ \Delta KE + \Delta PE = 0 \]
Substituting the expressions we derived:
\[ \frac{1}{2}mv_{\text{top}}^2 - \frac{1}{2}mv_{\text{min}}^2 + 2mgR = 0 \]
Substituting \( v_{\text{top}}^2 = gR \) into the equation:
\[ \frac{1}{2}m(gR) - \frac{1}{2}mv_{\text{min}}^2 + 2mgR = 0 \]
This simplifies to:
\[ \frac{1}{2}mgR - \frac{1}{2}mv_{\text{min}}^2 + 2mgR = 0 \]

### Step 5: Solve for \( v_{\text{min}}^2 \)
Combining terms, we get:
\[ \frac{1}{2}mgR + 2mgR = \frac{1}{2}mv_{\text{min}}^2 \]
\[ \frac{5}{2}mgR = \frac{1}{2}mv_{\text{min}}^2 \]
Dividing both sides by \( \frac{1}{2}m \):
\[ 5gR = v_{\text{min}}^2 \]
Thus, we find:
\[ v_{\text{min}} = \sqrt{5gR} \]

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