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
To solve this problem, we need to analyze the motion of a body initially at rest sliding down a frictionless track from a height \(h\) and then completing a vertical loop.
The problem requires us to find the minimum height \(h\) such that the body completes a vertical loop of diameter \(d\).
Concepts Involved:
- Conservation of Energy: The potential energy at height \(h\) is converted into kinetic energy at the lowest point of the loop.
- Centripetal Force Requirement: At the top of the loop, the body must have enough speed to maintain contact. The gravitational force and the required centripetal force must be balanced.
Step-by-Step Solution:
- At the top of the loop, the body must have enough kinetic energy to provide the centripetal force required for circular motion. The forces give us: \(mg = \frac{mv^2}{r}\), where \(r = \frac{d}{2}\).
- Solving for \(v\) at the top, we have \(v^2 = rg\).
- Using conservation of energy, the potential energy at height \(h\) is \(mgh\), and it converts into kinetic energy at the top of the loop, which includes both potential energy and kinetic energy: \(mgh = mg(2r) + \frac{1}{2}mv^2\).
- Substituting \(v^2 = rg\) into the energy conservation equation: \(mgh = mgd + \frac{1}{2}mrg\).
- Simplify to find \(h\): \(h = \frac{5}{4}d\).
Conclusion:
The minimum height \(h\) from which the body should start to just make the complete loop is \(\frac{5}{4}d\).
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