Question

A body initially at rest and sliding along a frictionless track from a height $h$ (as shown in the figure) just completes a vertical circle of diameter $AB \,=\, D$. The height h is equal to

• A. $\frac{5}{4} D $
• B. $\frac{3}{2} D $
• C. $\frac{7}{5} D $
• D. $D$

Answer 1

The Correct Option is A

To solve this problem, we need to determine the initial height \( h \) from which a body must be released so that it just completes a vertical circle of diameter \( D \). We'll consider the principles of energy conservation and dynamics of circular motion.

Step-by-step solution:

1. Initial Conditions: The body starts from rest at height \( h \). This implies its initial kinetic energy is zero, and its potential energy is \( mgh \) at this point.
2. At the top of the circle: To ensure the body just completes the circle, it needs just enough speed to stay on the track at the topmost point of the circle. This is a key condition for just completing the loop. At the top of the circle (height \( 2R = D \)), the minimum speed needed can be derived using the centripetal force condition: \[ v^2 = gR \] where \( R = \frac{D}{2} \). Hence, \[ v^2 = g \left(\frac{D}{2}\right) \]
3. Application of Conservation of Mechanical Energy: The total mechanical energy at the starting point is equal to that at the top of the loop: \[ mgh = mg(2R) + \frac{1}{2}mv^2 \] Substituting \( v^2 = gR \) into the equation and simplifying: \[ mgh = mgD + \frac{1}{2}mg \left(\frac{D}{2}\right) \] \[ mgh = mgD + \frac{1}{4}mgD \] \[ mgh = \frac{5}{4}mgD \] \[ h = \frac{5}{4}D \]
4. Thus, the required height \( h \) from which the body should be released to complete the vertical circle is \( \frac{5}{4} D \).

Hence, the correct answer is \( \frac{5}{4} D \).