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. \(\frac{3}{2}D\)
• B. \(\frac{7}{5}D\)
• C. D
• D. \(\frac{5}{4}D\)

Answer 1

The Correct Option is D

To determine the height \( h \) from which a body must be released to just complete a vertical circle, we can use the principles of energy conservation and the conditions necessary for an object to perform circular motion.

Given:
- Diameter of the vertical circle, \( AB = D \).
- Therefore, the radius \( R = \frac{D}{2} \).

Step 1: Energy Conservation

At the top of the loop, the body must have enough kinetic energy to maintain motion against gravity. By energy conservation, the initial potential energy is converted into kinetic energy at the top of the loop.

Initial potential energy at height \( h \):
\[ PE_i = mgh \] Potential energy at the top of the circle (height \( 2R \)):
\[ PE_f = mg(2R) = mgD \] Kinetic energy required at the top of the loop:
To ensure the body just completes the circle, the centripetal force at the top: \[ \frac{mv^2}{R} = mg \implies v^2 = gR \] Kinetic energy at the top:
\[ KE = \frac{1}{2}mv^2 = \frac{1}{2}mgR \]

Step 2: Total Energy at the Top of the Loop

The total energy at the top of the loop is the sum of kinetic and potential energies:
\[ PE_f + KE = mgD + \frac{1}{2}mgR = mgD + \frac{1}{2}mg\frac{D}{2} = mgD + \frac{1}{4}mgD = \frac{5}{4}mgD \]

Step 3: Equate Initial and Final Energies

By energy conservation, the initial potential energy equals the total energy at the top:
\[ mgh = \frac{5}{4}mgD \] Cancel \( mg \) from both sides:
\[ h = \frac{5}{4}D \]

Thus, the initial height \( h \) from which the body must be released is \( \frac{5}{4}D \).

The correct answer is: \(\frac{5}{4}D\).