Planar rigid bar kinematics (instantaneous angular motion): A 5 m rod moves in a vertical plane. When the rod is inclined at 60° to the horizontal, its lower end moves horizontally at 3 m/s, while the upper end moves purely vertically. What is the speed of the upper end (approximate to the nearest option)?

Introduction / Context:This is a classic rigid-body planar kinematics problem. A straight bar undergoing pure rotation at an instant has point velocities related by the cross product of angular velocity with position vectors. When one end has a known velocity and the other end is constrained along a perpendicular direction, we can solve for the instantaneous angular speed and the unknown end speed.

Given Data / Assumptions:

• Rod length L = 5 m; rod makes 60° with the horizontal.
• Lower end A velocity: v_A = 3 m/s horizontally.
• Upper end B moves vertically (its instantaneous horizontal component is zero).
• Planar motion; rigid bar constraint; no deformation.

Concept / Approach:Use velocity relation v_B = v_A + ω × r_BA. With r_BA = (L cos60°, L sin60°) = (2.5, 4.330) m from A to B, and ω about the out-of-plane axis, we write component equations and solve for ω and vertical speed at B.

Step-by-Step Solution:

Verification / Alternative check:An equivalent constraint approach demands equal components of the two-end velocities along the rod (no change in length), which also yields v_B ≈ 1.73 m/s. The computed value rounds to the nearest listed option.

Why Other Options Are Wrong:

• 0.5, 1.0 m/s: too small; would imply unrealistically low ω for the given 3 m/s at the lower end.
• 2.5, 3.0 m/s: too large; would require a higher ω inconsistent with B’s zero x-velocity constraint.

Common Pitfalls:Forgetting that v_Bx must be zero; or using L instead of its components (x, y) with the angle; or assuming pure translation.

Final Answer:1.5 m/s