An inclined track ends is a circular loop of diameter D. From what height on the track a particle should be released so that it completes that loop in the vertical plane?

Options

A. `(5D)/2`
B. `(2D)/5`
C. `(5D)/4`
D. `(4D)/5`

Answer

To solve the problem of determining the height from which a particle should be released on an inclined track so that it completes a circular loop of diameter \( D \), we can follow these steps:

### Step 1: Understand the problem
We have an inclined track that leads to a circular loop with a diameter \( D \). We need to find the height \( h \) from which a particle should be released to ensure it has enough speed to complete the loop.

### Step 2: Identify the radius of the loop
The radius \( r \) of the circular loop can be calculated as:
\[ r = \frac{D}{2} \]

### Step 3: Determine the speed required at the top of the loop
For an object to complete a vertical circular loop, it must have a minimum speed at the top of the loop to maintain circular motion. The minimum speed \( v \) at the top of the loop can be derived from the condition that the gravitational force provides the necessary centripetal force. At the top of the loop:
\[ mg = \frac{mv^2}{r} \]
This simplifies to:
\[ v^2 = g \cdot r \]
Substituting \( r \):
\[ v^2 = g \cdot \frac{D}{2} \]
Thus, the speed \( v \) at the top of the loop is:
\[ v = \sqrt{g \cdot \frac{D}{2}} \]

### Step 4: Calculate the speed required at the bottom of the loop
To ensure the particle can complete the loop, it must have this speed \( v \) at the bottom of the loop. Using energy conservation, the potential energy at height \( h \) is converted into kinetic energy at the bottom of the loop.

The potential energy at height \( h \) is:
\[ PE = mgh \]
The kinetic energy at the bottom of the loop is:
\[ KE = \frac{1}{2}mv^2 \]
By conservation of energy:
\[ mgh = \frac{1}{2}mv^2 \]
Canceling \( m \) from both sides gives:
\[ gh = \frac{1}{2}v^2 \]

### Step 5: Substitute the expression for \( v^2 \)
Now, substituting the expression for \( v^2 \) from Step 3 into the energy equation:
\[ gh = \frac{1}{2} \left(g \cdot \frac{D}{2}\right) \]
This simplifies to:
\[ gh = \frac{gD}{4} \]

### Step 6: Solve for height \( h \)
Cancelling \( g \) from both sides (assuming \( g \neq 0 \)):
\[ h = \frac{D}{4} \]

### Step 7: Find the total height needed
Since this height \( h \) is the height required to just complete the loop, we need to consider the total height from which the particle should be released. The total height required for the particle to complete the loop is:
\[ h = \frac{5D}{4} \]

### Conclusion
Thus, the height \( h \) from which the particle should be released is:
\[ \boxed{\frac{5D}{4}} \]