A particle of mass $m$ is attached to a light string of length $L$, which is fixed at one end. The particle is released from rest when the string is horizontal. At the lowest point, the string hits a small peg located at a distance $d$ below the fixed point. What is the minimum value of $d$ such that the particle completes a full vertical circle around the peg?

Solution

1. By conservation of energy, the velocity at the lowest point is $v=\sqrt{2gL}$.
2. For a circle of radius $r=L-d$, the velocity at the top must satisfy $v_{top}^2\ge g(L-d)$.
3. Using energy conservation from the bottom to the top of the small circle, $\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_{top}^2+mg(2(L-d))$.
4. Substituting $v^2=2gL$ and $v_{top}^2=g(L-d)$, we get $2gL = g(L - d) + 4g(L - d) = 5g(L - d)$.
5. Solving $2L = 5L - 5d$ gives $5d = 3L$, so $d=0.6L$.
6. Hence the answer is (C).