A 2-lb block rests on the smooth semicylindrical surface at A. An elastic cord having a stiffness of k = 2 lb/ft is attached to the block at B and to the base of the semicylinder at C. If the block is released from rest at A, determine the longest unstretched length of the cord so the block begins

Question

A 2-lb block rests on the smooth semicylindrical surface at A. An elastic cord having a stiffness of k = 2 lb/ft is attached to the block at B and to the base of the semicylinder 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 heta = 45°. Neglect the size of the block.

Accepted Answer

Equation of Motion: It is required that N = 0. Applying the bellow, we have

[latex]\Sigma { F }_{ n }={ ma }_{ n };\quad \quad 2\cos { { 45 }^{ \circ } } =\frac { 2 }{ 32.2 } (\frac { { v }^{ 2 } }{ 1.5 } )\quad \quad { v }^{ 2 }=34.15{ m }^{ 2 }/{ s }^{ 2 }[/latex]

Potential Energy: Datum is set at the base of cylinder. When the block moves to a position [latex] 1.5 \sin { { 45 }^{ \circ } } = 1.061 ft[/latex] above the datum, its gravitational potential energy at this position is [latex] 2(1.061) = 2.121 ft \cdot lb[/latex]. The initial and final elastic potential energy are [latex] \frac { 1 }{ 2 } (2)[\pi (1.5)-l]^{ 2 }[/latex] and [latex] \frac { 1 }{ 2 } (2)[0.75\pi (1.5)-l]^{ 2 }[/latex], respectively.

Conservation of Energy:

[latex] \Sigma { T }_{ 1 }+\Sigma { V }_{ 1 }=\Sigma { T }_{ 2 }+\Sigma { V }_{ 2 }\ 0+\frac { 1 }{ 2 } (2)[\pi (1.5)-l]^{ 2 }=\frac { 1 }{ 2 } (\frac { 2 }{ 32.2 } )(34.15)+2.121+\frac { 1 }{ 2 } (2)-[0.75\pi (1.5)-l]^{ 2 }\ l = 2.77 ft[/latex]

Question : A 2-lb block rests on the smooth semicylindrical surface at ...