A 2 lb block rests on the smooth cylinderical surface at A. An elastic cord having a stiffness of k = 2lb/ft is attached to the block at B and the base of the cylinder at C. If the block is released from rest at A, determine the longest unstretched length of the cord so the block begins to leave the cylinder at the instant
q = 45^o. Neglect the size of the block.

1) 6.9 ft
2) 5.5 ft
3) 4.5 ft
4) 7.5 ft

Equation of motion:
It is required that N = 0. Summing up all the forces in the normal direction we have
S Fn = ma_n and 2cos45^o = 2/32.2(v²/3)
v² = 68.3 m²/s²

Potential energy: Datum is at the base of cylinder. When the block moves to a position
3sin45^o = 4.24 fl.lb
The initial and the final elastic potential energy are
1/2 (2) [p(3) - l]² and
1/2 (2) [0.75p(3) - l]²

Applying the conservation of energy principle we have:
ST₁ + S V₁ = ST₂ + S V₂
1/2 (2) [p(3) - l]² = 1/2(2/32.2)(68.3) + 4.24 + 1/2 (2) [0.75p(3) - l]²
1.5 p l = 32.05
l = 21.67/p = 6.9 ft

So the longest unstretched lenght of the cord so that the block begins to leave the cylinder at the instant of 45^o is l = 6.9 ft