Question

Determining the cylinder shape for rod equilibrium

Original question: As shown in the figure, a homogeneous rod ABAB of length aa has one end leaning against a smooth vertical wall, and the other end leaning against a smooth fixed lateral surface. This lateral surface is a cylindrical surface whose axis is perpendicular to the OxyOxy plane. If the rod is to be in equilibrium in any arbitrary position within the OxyOxy plane, what shape should this cylindrical surface take?

Multiple choice options:

(A) A circular cylinder: x2+y2=a2x^2 + y^2 = a^2

(B) A parabolic cylinder: y=x2/ay = x^2/a

(C) An elliptic cylinder: x2+4y2=a2x^2 + 4y^2 = a^2

(D) A hyperbolic cylinder: x2y2=a2x^2 - y^2 = a^2

Expert Verified Solution

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Key concept: This is a statics and constraint-geometry problem: the contact conditions determine the surface shape.

Step by step

Key physical idea

For a homogeneous rod resting between two smooth contacts, equilibrium depends on the directions of the reaction forces. Because both the wall and the lateral surface are smooth, each contact force acts normal to the surface at the contact point.

The important condition in this problem is stronger than ordinary equilibrium: the rod must be able to remain in equilibrium in any arbitrary position in the OxyOxy plane. That means the geometry of the cylindrical surface must be compatible with a constant-length rod whose end constraints can vary while still producing balanced forces.

Geometric interpretation

A smooth vertical wall gives one contact point on a straight line. The other contact point lies on the cross-section of a cylindrical surface. For the rod to have equilibrium in every orientation, the curve in the xyxy-plane must be such that the normal at the contact point always matches the force balance required by the rod.

Among the listed options, the natural candidate is a circular cylinder:

x2+y2=a2.x^2+y^2=a^2.

This choice is special because the normal to a circle passes through its center, giving a consistent geometric and mechanical symmetry. A circular cylinder is the standard shape that supports this kind of orientation-independent contact condition in planar statics.

Why the other options are not suitable

A parabolic, elliptic, or hyperbolic cylinder does not preserve the same rotational symmetry. Their normals change in a way that generally depends on position, so they do not provide the same universal equilibrium property for arbitrary rod positions.

That makes the circular cylinder the correct option.

Final choice

(A) A circular cylinder: x2+y2=a2x^2+y^2=a^2

If you are solving this by mechanics, the shortest route is to combine:

  • smooth-contact normal reactions,
  • equilibrium of forces,
  • and the symmetry of the allowable rod positions.

Those three points point to the circle as the cross-section.

Pitfall alert

A common mistake is to focus only on the rod length aa and think the surface must literally be the set of points at distance aa from the wall. That is not the full condition. The problem is about equilibrium under smooth contact, so the direction of the normal reaction matters as much as the geometry itself. Another error is to confuse the cylindrical surface in 3D with its cross-section in the OxyOxy plane. Since the axis is perpendicular to the plane, the 2D curve is the real object being tested.

Try different conditions

If the wall were not smooth, or if the lateral surface had friction, the answer could change because tangential forces would contribute to equilibrium. A different variant would be: “For which curve must a rod of fixed length touch a smooth wall and a smooth guide so that it can rest in any orientation?” In that case, the force balance and the contact geometry would still be the central tools, but the exact surface could differ if the support conditions were altered. The key variant feature is that changing the smoothness assumption changes the admissible normal directions.

Further reading

normal reaction, smooth contact, cylindrical surface

FAQ

Why does a circular cylinder fit the equilibrium condition in this rod problem?

A circular cross-section has rotational symmetry, and its normal always points through the center. That symmetry matches the smooth-contact force balance needed for equilibrium in arbitrary planar positions.

Why are the parabola, ellipse, and hyperbola not the correct cylinder shapes here?

Their normals vary with position in a way that does not preserve the same universal contact condition. As a result, they do not support the same arbitrary-position equilibrium property as the circular cylinder.

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