A body initially at rest and sliding along a frictionless track from a height 'h' (as shown in figure) just completes a vertical circle of diameter AB = d. The height 'h' is equal to
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Always pay attention to whether the question gives the radius \( R \) or the diameter \( d \). A common mistake is using the formula \( h = \frac{5}{2}R \) and selecting an option assuming \( R = d \text{ (diameter)} \), which leads to incorrect answers.
Step 1: Understanding the Question:
A body slides down from a height \( h \) on a frictionless track and enters a vertical circular loop of diameter \( d \). We need to determine the minimum height \( h \) required so that the body can just complete the vertical circle. Step 2: Key Formula or Approach:
1. Conservation of Mechanical Energy: Since the track is frictionless, the initial potential energy at height \( h \) is completely converted into kinetic energy at the lowest point of the circular track (point B):
\[ mgh = \frac{1}{2}mv^2 \]
2. Condition to complete a vertical circle: The minimum velocity \( v \) required at the lowest point of a vertical circle of radius \( R \) to just complete the loop is:
\[ v_{\text{min}} = \sqrt{5gR} \] Step 3: Detailed Explanation:
Let the radius of the circular track be \( R \). The diameter of the circle is given as:
\[ d = 2R \implies R = \frac{d}{2} \]
Using the conservation of energy between the starting point and the lowest point:
\[ mgh = \frac{1}{2}m v^2 \]
To just complete the circle, the velocity at the lowest point must be at least \( v_{\text{min}} = \sqrt{5gR} \). Substituting this value:
\[ mgh = \frac{1}{2}m (\sqrt{5gR})^2 \]
\[ mgh = \frac{5}{2}mgR \]
Canceling \( mg \) on both sides:
\[ h = \frac{5}{2}R \]
Now, substitute \( R = \frac{d}{2} \) into the equation:
\[ h = \frac{5}{2}\left(\frac{d}{2}\right) = \frac{5}{4}d \] Step 4: Final Answer:
The minimum height \( h \) is \( \frac{5}{4}d \).
