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 Concept:
This problem involves the conservation of mechanical energy and the conditions for motion in a vertical circle.
A body sliding down a frictionless track converts gravitational potential energy into kinetic energy.
To "just complete" a vertical circle, the body must have a minimum critical velocity at the lowest point of the circle.
At the highest point of the circle, the tension in the string (or normal force) just becomes zero, and gravity provides the necessary centripetal force. Step 2: Key Formula or Approach:
Let \(R\) be the radius of the circle.
Condition to complete the circle: Minimum velocity at the bottom \(v_{bottom} = \sqrt{5gR}\).
Relationship between diameter \(d\) and radius \(R\): \(d = 2R \implies R = d/2\).
Conservation of Energy: \(mgh = \frac{1}{2} m v_{bottom}^2\). Step 3: Detailed Explanation:
The diameter of the circle is \(d\). Therefore, the radius is:
\[ R = \frac{d}{2} \]
For an object to just complete a vertical circle of radius \(R\), the minimum velocity required at the lowest point (point B in the figure) is:
\[ v_B = \sqrt{5gR} \]
The kinetic energy at the bottom must be:
\[ K.E._B = \frac{1}{2} m v_B^2 = \frac{1}{2} m (5gR) = \frac{5}{2} mgR \]
Initially, the body is at rest at height \(h\). Its initial mechanical energy is purely potential:
\[ P.E._{initial} = mgh \]
By the Law of Conservation of Energy (since the track is frictionless):
\[ P.E._{initial} = K.E._B \]
\[ mgh = \frac{5}{2} mgR \]
Cancel \(mg\) from both sides:
\[ h = \frac{5}{2} R \]
Now, substitute \(R = d/2\) to express the result in terms of the diameter:
\[ h = \frac{5}{2} \left( \frac{d}{2} \right) \]
\[ h = \frac{5}{4} d \]
This matches option (B). Step 4: Final Answer:
The required height for a body to just complete a vertical circle is \(2.5\) times the radius. In terms of diameter, this corresponds to \(5/4\) of the diameter.
