University Physics I · 2D & 3D Kinematics · 3.7
Kinematic constraints
Connected objects do not move independently. The geometry that links them — a taut rope, a touching surface, a wheel that does not slip — is an equation, and its time derivatives are the kinematics.
Build the model
Connect the measurement to the mechanism.
A constraint is a geometric fact that stays true throughout the motion: a rope keeps its length, touching surfaces neither overlap nor separate, a rolling rim does not slide. Write it as an equation between coordinates measured from fixed points, then differentiate — once for velocities, twice for accelerations.
The signs come from your coordinates, not from intuition. No force is involved, and each independent constraint removes one unknown.
- Simple definition
- A kinematic constraint is a geometric condition connected objects satisfy at every instant, so differentiating it relates their velocities and then their accelerations.
- Example
- Two blocks on a taut rope over a fixed pulley, each coordinate measured from the pulley: x₁ + x₂ = L, so v₁ = −v₂ and a₁ = −a₂.
Measure every coordinate from a fixed point, or the derivatives lose their meaning.
positions in m → velocities in m s⁻¹ → accelerations in m s⁻²
The two blocks share one speed and one acceleration magnitude while the rope stays taut.
x₁, x₂ measured from the pulley along each segment · L is the rope length less the fixed arc, in m
Free end taken in at 0.60 m s⁻¹ lifts the load at 0.30 m s⁻¹.
all coordinates measured down from the ceiling, in m
h = 4.0 m, x = 3.0 m, reel-in 1.5 m s⁻¹ → boat speed 2.5 m s⁻¹.
x horizontal distance · h fixed winch height · ℓ taut rope length, in m · θ between rope and velocity
Foot out at 0.40 m s⁻¹ with x = 3.0 m, y = 4.0 m drops the top at 0.30 m s⁻¹.
n̂ along the rod, or along the common normal at a contact · leaning rod x² + y² = L² gives x vₓ + y vᵧ = 0
R = 0.35 m with ω = 12 rad s⁻¹: centre 4.2 m s⁻¹, rim top 8.4 m s⁻¹.
ω = dθ/dt in rad s⁻¹ · α = dω/dt in rad s⁻² · s and R in m · vcm in m s⁻¹ · acm in m s⁻²
A constraint is an equation about geometry
A constraint is a geometric statement that stays true for the whole motion: the rope keeps its length, the surfaces stay in contact, the rim does not slide. Write it as an equation between coordinates measured from fixed points — an origin that rides on one of the moving objects gives derivatives that do not mean what you want them to. After that the kinematics is calculus: differentiate once for a relation between velocities, twice for a relation between accelerations. Force never enters. The counting is the payoff. Two objects free to move along a line carry two unknown accelerations; one rope between them supplies one equation, leaving a single independent motion for Newton's second law to determine later.
Ropes: the length is what stays constant
An ideal rope is inextensible and stays taut, so its two straight segments add to a constant L — the rope length less the fixed arc over the pulley. With x₁ and x₂ measured from the pulley along each segment, x₁ + x₂ = L. Differentiating gives v₁ + v₂ = 0, that is v₁ = −v₂, and then a₁ = −a₂: whatever one block gains, the other loses, in those coordinates. That minus sign is bookkeeping rather than physics. It follows from measuring both coordinates away from the same point, and it flips if one coordinate is measured the other way. The equation dies the instant the rope goes slack or stretches, so state that idealisation when you use it.
Movable pulleys: count the supporting segments
When a pulley itself moves, count how many rope segments run to it. Fix one end of the rope to the ceiling, pass it under a movable pulley carrying the load, then back up over a fixed pulley to your hand: the load hangs on two segments, so every metre you take in is shared between them and the load rises half a metre. With every coordinate measured down from the ceiling, 2yₚ + yfree = L, so vₚ = −½ vfree and aₚ = −½ afree. Take the free end in at 0.60 m s⁻¹ and the load rises at 0.30 m s⁻¹. Thread the same rope a different way and the ratio changes, so re-derive the constraint for each arrangement instead of remembering a factor.
At an angle, only the component along the rope counts
A rope is not always parallel to the motion it controls. A winch a height h above the water reels in a boat at horizontal distance x, so ℓ² = x² + h², and because h is fixed, ℓ(dℓ/dt) = x(dx/dt). Written another way, v cos θ = |dℓ/dt| with θ the angle between rope and velocity: the rope shortens at the rate of the velocity component along it, so the boat outruns the reel-in rate by a factor 1/cos θ. With h = 4.0 m, x = 3.0 m and a reel-in rate of 1.5 m s⁻¹, the rope is 5.0 m long and the boat moves at 2.5 m s⁻¹. As x → 0 the predicted speed diverges, which is where the straight-taut-rope model fails, not the boat.
Contact and rigid links match components along one line
Two bodies that stay in contact without overlapping or separating must have equal velocity components along their common normal. The tangential components may differ, and that difference is the sliding. A rigid rod is the same statement with the line taken along the rod, so vA · n̂ = vB · n̂ for its two ends. Take a 5.0 m rod with its foot on the floor and its top against a wall: x² + y² = L² gives x vₓ + y vᵧ = 0, so a foot sliding out at 0.40 m s⁻¹ when x = 3.0 m and y = 4.0 m sends the top down at 0.30 m s⁻¹. A block on a moving wedge obeys the same rule along the incline's normal: its velocity relative to the wedge stays parallel to the surface.
Rolling ties a translation to a rotation
A wheel rolling without slipping lays its rim onto the ground, so the arc Rθ equals the distance the centre travels: s = Rθ. Differentiating that one geometric fact gives vcm = ωR and acm = αR, with ω = dθ/dt and α = dω/dt. Read at the contact, the same condition says that point is instantaneously at rest, which puts the centre at ωR and the top of the rim at 2ωR. With R = 0.35 m and ω = 12 rad s⁻¹ the centre moves at 4.2 m s⁻¹ and the top at 8.4 m s⁻¹; α = 8.0 rad s⁻² gives acm = 2.8 m s⁻². If the surface moves as well, the condition applies to the relative velocity there: vcm − vsurface = ωR. For a wheel confined to one plane this relation integrates back to a relation between positions; for a ball free to roll over a plane it does not, and the constraint restricts velocities only.
Change one variable at a time
Make the relationship visible.
Drag the boat in toward the winch with the reel-in rate held fixed and watch the horizontal arrow outgrow the one along the rope — the dashed line is the perpendicular foot, so the rope arrow is exactly v cos θ.
ROPE LENGTH ℓ6.40 m
ROPE ANGLE θ38.7 °
cos θ0.781
BOAT SPEED v1.92 m s⁻¹
Live interpretationROPE LENGTH ℓ: 6.40 m. ROPE ANGLE θ: 38.7 °. cos θ: 0.781. BOAT SPEED v: 1.92 m s⁻¹
Catch the common trap
Explain before calculating.
A wheel of radius 0.25 m rolls without slipping on a conveyor belt whose surface moves forward at 1.2 m s⁻¹. The wheel's centre moves forward at 2.0 m s⁻¹. What is the wheel's angular speed?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA rope is tied to the ceiling, runs down under a movable pulley carrying a crate, back up over a fixed pulley, and down to your hands. You take the free end in at 0.90 m s⁻¹ at t = 0, steadily increasing to 1.50 m s⁻¹ at t = 2.0 s. Find the crate's speed and acceleration at t = 2.0 s.
- Measure every coordinate down from the ceiling. Two segments run down to the movable pulley and one runs down to your hands, so 2yₚ + yfree = L.
- Differentiate once: 2vₚ + vfree = 0, so vₚ = −½vfree. Pulling the free end down is vfree = +1.50 m s⁻¹, giving vₚ = −0.75 m s⁻¹ — negative is upward in these coordinates.
- Differentiate again: 2aₚ + afree = 0. The pull rate climbs steadily, so afree = (1.50 − 0.90) ÷ 2.0 = 0.30 m s⁻², and aₚ = −0.15 m s⁻².
- Both halvings come from the crate hanging on two segments. Rethread the rope and the factor changes, so the geometry has to be rewritten, not remembered.
AnswerThe crate rises at 0.75 m s⁻¹ and is accelerating upward at 0.15 m s⁻².
MediumA 5.0 m ladder leans against a vertical wall with its foot on the floor. The foot is dragged away from the wall at a constant 0.60 m s⁻¹. When the foot is 3.0 m from the wall, find the velocity and the acceleration of the top.
- The ladder is rigid, so its ends satisfy x² + y² = L² = 25 m² at every instant, with x measured from the wall and y up the wall.
- At x = 3.0 m: y = √(25 − 9.0) = 4.0 m.
- Differentiate once: x vₓ + y vᵧ = 0, so vᵧ = −(3.0 × 0.60) ÷ 4.0 = −0.45 m s⁻¹ — the top slides down at 0.45 m s⁻¹.
- Differentiate again: vₓ² + x aₓ + vᵧ² + y aᵧ = 0. The foot moves at constant velocity, so aₓ = 0 and aᵧ = −(0.60² + 0.45²) ÷ 4.0 = −0.5625 ÷ 4.0 = −0.14 m s⁻².
- A constant-speed foot does not give a constant-speed top: the constraint is quadratic, so differentiating it twice leaves a term in the velocities that nothing cancels.
AnswerThe top moves down at 0.45 m s⁻¹ and accelerates downward at 0.14 m s⁻².
HardA cable is wound round the rim of a wheel of radius R = 0.30 m standing on the ground. The cable leaves the rim at the very top and runs horizontally to a winch that reels it in at a steady 2.4 m s⁻¹. The wheel rolls without slipping. Find the speed of the centre, the angular speed, and the length of cable taken in while the wheel advances 3.0 m.
- No slip means the contact point is instantaneously at rest, so the wheel turns about it: the centre, a height R above the contact, moves at vcm = ωR, and the top of the rim, a height 2R above it, moves at 2ωR = 2vcm.
- The cable does not slip on the rim, so its straight run is drawn in at the speed of the point where it leaves — the top. Hence 2vcm = 2.4 m s⁻¹ and vcm = 1.2 m s⁻¹.
- ω = vcm ÷ R = 1.2 ÷ 0.30 = 4.0 rad s⁻¹.
- For a wheel confined to one plane the rolling relation integrates back to a relation between positions, so the 2:1 speed ratio is also a 2:1 length ratio: 2 × 3.0 = 6.0 m of cable per 3.0 m of travel.
Answervcm = 1.2 m s⁻¹, ω = 4.0 rad s⁻¹, and 6.0 m of cable for 3.0 m of travel.