An inclined track ends in a circular loop of radius r. From what height on the track a particle should be released so that it completes the loop, assuming there is no friction ?

Options

A. `r/2`
B. `(3r)/2`
C. `2r`
D. `(5r)/2`

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 radius \( r \), we can follow these steps:

### Step 1: Understand the conditions for completing the loop
To complete the circular loop, the particle must have a minimum velocity at the top of the loop. This minimum velocity \( v \) can be derived from the centripetal force requirement. At the top of the loop, the gravitational force must provide the necessary centripetal force:

\[ mg = \frac{mv^2}{r} \]

From this, we can derive the minimum velocity at the top of the loop:

\[ v^2 = g r \quad \Rightarrow \quad v = \sqrt{g r} \]

However, this is the condition for the top of the loop. For the particle to just complete the loop, we actually need to consider the energy conservation from the starting height to the top of the loop.

### Step 2: Apply conservation of energy
When the particle is released from height \( h \), its initial potential energy is converted into kinetic energy and potential energy at the top of the loop.

The potential energy at height \( h \) is:

\[ PE_{\text{initial}} = mgh \]

At the top of the loop (height \( 2r \)), the potential energy is:

\[ PE_{\text{top}} = mg(2r) \]

The kinetic energy at the top of the loop is:

\[ KE_{\text{top}} = \frac{1}{2} mv^2 \]

Using the minimum velocity \( v = \sqrt{5gr} \) at the top of the loop (derived from the centripetal force condition), we can substitute this into the kinetic energy expression:

\[ KE_{\text{top}} = \frac{1}{2} m(5gr) = \frac{5}{2} mgr \]

### Step 3: Set up the energy conservation equation
By conservation of energy, we have:

\[ mgh = PE_{\text{top}} + KE_{\text{top}} \]

Substituting the expressions we derived:

\[ mgh = mg(2r) + \frac{5}{2} mgr \]

### Step 4: Simplify the equation
Dividing through by \( mg \) (assuming \( m \neq 0 \)) gives:

\[ h = 2r + \frac{5}{2}r \]

Combining the terms:

\[ h = 2r + 2.5r = 4.5r = \frac{9r}{2} \]

### Step 5: Conclusion
Thus, the height \( h \) from which the particle should be released is:

\[ h = \frac{5r}{2} \]

### Final Answer
The correct height from which the particle should be released to complete the loop is:

\[ h = \frac{5r}{2} \]