WE-6 — Worked Examples: Constraints Without Strings

Mahavakya

A constraint is a geometric fact that does not change while everything else does. Find it, write it as an equation, and the kinematics is differentiation.

The six, and what is held fixed

1. The sliding ladder  a length
2. A rod riding a moving wedge  an angle
3. The scissor lift  a link length
4. Crank and piston  two lengths
5. A rope round a moving pin  a total length
6. A bead on a spinning hoop  a radius

Problem

A 5 m ladder leans against a vertical wall. Its base slides away from the wall at a steady 2 m/s.

When the base is 3 m from the wall, find the speed and the acceleration of the top end.

x² + y² = 25

differentiate:   xv_x + yv_y = 0

v_x² + xa_x + v_y² + ya_y = 0

The base moves steadily, so a_x = 0:

4 + 2.25 + 4a_y = 0  →  a_y = −1.56 m/s²

The base has zero acceleration and the top does not. Nothing is pushing the top harder as time passes — the geometry alone forces it to accelerate, because as the ladder flattens, the same base speed demands an ever-faster drop.

Push it further: as y → 0 the relation xv_x = −yv_y forces v_y → ∞. The model breaks before the ladder does — a real ladder leaves the wall first, which the constraint does not know about.

Problem

A wedge of face angle θ slides horizontally at speed V. A rod, held so that it can only move vertically, rests on the sloping face.

Find the rod’s speed.

y = (X₀ − X) tan θ

dy/dt = −(dX/dt) tan θ

with dX/dt = −V:   v_y = V tan θ

This is the only constraint in the module whose factor is unbounded. A string gives a ratio of 2, or 3, or 1/2. Here the ratio is tan θ, and as θ → 90° it grows without limit — a nearly vertical face lifts the rod arbitrarily fast for any wedge speed you like.

Compare it with the wedge in the first set, where a block slides along the face. Here the rod cannot — it is a follower, and the relation is different because the permitted motion is different.

Problem

A two-stage scissor lift is built from links of length b = 1.5 m. An actuator draws the base pins together at 0.8 m/s.

Find the speed of the platform when the links make 45° with the horizontal.

x = 2b cos θ     H = 4b sin θ

Eliminate dθ/dt between the two derivatives:

v_H = −2 cot θ · (dx/dt)

= −2(1)(−0.8) = +1.6 m/s, rising

The angle is the hinge of the problem, and it cancels. Neither x nor H alone gives the answer — but each is a function of θ, so dθ/dt is the bridge between them and disappears on the way through.

Notice also that b vanishes. The link length does not affect the speed ratio, only the reach — which is why lifts of very different sizes handle the same way.

Problem

A crank of radius r = 0.1 m turns at a constant 100 rad/s. A connecting rod of length l = 0.3 m links the crank pin to a piston sliding in a horizontal cylinder.

Find the piston’s velocity when the crank is at 90°.

x_Q = r cos θ + √(l² − r² sin² θ)

v_Q = −ωr[ sin θ + r sin 2θ / (2√(l² − r² sin² θ)) ]

v_Q = −(100)(0.1) = −10 m/s

At 90° the piston moves at exactly ωr, as if the rod were infinitely long. The rod’s contribution is carried entirely by sin 2θ, which vanishes at 0°, 90° and 180° — so at those three positions the messy square root never has to be evaluated.

That is worth noticing before differentiating anything. Substituting the angle first would have saved most of the algebra, and the general expression is only needed if a general answer is wanted.

Problem

A rope is anchored at the origin, passes around a pin Q sliding along the ground at constant speed u, then runs up over a fixed peg at height h directly above the anchor, and down to a hanging mass.

Find the mass’s speed when the pin is at horizontal position x.

L = x + √(x² + h²) + y_M

differentiate:   u + [x/√(x² + h²)]u + dy_M/dt = 0

v_M = −u [ 1 + x/√(x² + h²) ]

The bracket runs from 1 to 2. Near the anchor (x → 0) only the ground segment is lengthening and the mass falls at u. Far away (x ≫ h) the slanted segment has become nearly horizontal too, both grow at almost u, and the mass falls at nearly 2u.

A ratio that varies continuously with position — no fixed factor exists to memorise, which is the point.

Problem

A bead slides on a circular wire hoop of radius 0.5 m. The hoop spins about its vertical diameter at 6 rad/s. At the instant the bead is 60° from the bottom, it is sliding along the wire at dθ/dt = 2 rad/s.

Find the bead’s speed in space.

along the wire:   v_θ = R(dθ/dt) = 0.5(2) = 1.0 m/s

carried by the spin:   v_φ = (R sin θ)Ω = 0.5(0.866)(6) = 2.60 m/s

|v| = √(1.0² + 2.60²) = 2.78 m/s

The spin term uses R sin θ, not R. The bead circles the vertical axis on a ring of radius R sin θ — which is zero at the bottom of the hoop and largest at the equator. A bead sitting at the very bottom is carried nowhere at all by the rotation, however fast the hoop spins.

The two motions are perpendicular, so they combine by Pythagoras — K15’s rule, in three dimensions this time.

What the six have in common

A ladder, a wedge, a lift, an engine, a rope and a hoop. Not one shares a mechanism with another, and every one was solved in the same three moves: name the thing that does not change, write it as an equation, differentiate.

The factors that came out could not have been recalled. Two of them — tan θ and the bracket in example 5 — are not constants at all, and the ladder’s is not even bounded. There was never a rule to remember, only a geometry to differentiate.

Which is why the first constraints set and this one need no separate methods. Strings were never the subject. Fixed lengths were.

Carried forward

• Ask what stays fixed before asking what moves.
• Differentiate once for velocities, twice for accelerations — and expect the product rule the second time.
• A constraint factor need not be constant. It can depend on position, and it need not be bounded.
• Where two quantities share an angle, that angle is the bridge and it usually cancels.
• Substituting a specific angle before differentiating often removes most of the algebra.

Prerequisites: K14, Foundation 8  ·  Working time: 45 min  ·  Level: JEE Advanced