Modelling and vectors in mechanics flashcards

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Particle (modelling)
An object whose mass is taken to be concentrated at a single point: its size, shape and rotation are ignored.
Rigid body
An object whose shape does not change: bending, stretching and squashing are ignored.
Rod (modelling)
A rigid body with length but negligible thickness — treated as one-dimensional.
Lamina
A flat object with area but negligible thickness.
Light (string, rod or pulley)
Its mass is negligible, so its weight is ignored; a light string has the same tension throughout.
Inextensible string
A string that does not stretch, so particles joined by it (while taut) have the same speed and the same size of acceleration.
Smooth surface
No friction acts; the only contact force is perpendicular to the surface.
Light smooth pulley
A pulley with no mass and no friction at its axle, so the tension in a string passing over it is the same on both sides.
Bead (modelling)
A particle with a hole through it, threaded on a wire or string, so it can only move along that wire or string.
Smooth peg
A support of negligible size with no friction; the force it exerts is perpendicular to the object resting on it.
Vector
A quantity with magnitude and direction, e.g. displacement, velocity, acceleration, force. Speed, distance and mass are scalars.
Unit vector
A vector of magnitude 1. The unit vector in the direction of v is v/|v|; i and j are perpendicular unit vectors.
Bearing
A direction measured clockwise from north, written with three figures (e.g. 059°).
Magnitude of xi + yj
|xi + yj| = √(x² + y²). (No — derive it)
Vector of magnitude F at angle θ (anticlockwise) to i
F cos θ i + F sin θ j. (No — derive it)
Vector of magnitude F on a bearing θ (i east, j north)
F sin θ i + F cos θ j. (No — derive it)
Position with constant velocity
r = r₀ + vt. (No — must be known)
Constant-acceleration formulae in vector form
v = u + at; r = r₀ + ut + ½at². (No — must be known)
Newton's second law in vector form
F = ma, where F is the resultant force; F and a are parallel. (No — must be known)
Vector diagram for a resultant
Draw the vectors tip to tail; the resultant runs from the first tail to the last tip.
Forces in equilibrium on a vector diagram
Drawn tip to tail they form a closed polygon (a closed triangle for three forces).
Shape of the path when a is not parallel to u (constant a)
A curve (part of a parabola) bending towards the direction of a.
Graph of distance between two constant-velocity objects against time
A curve with a single minimum (the square root of a quadratic); the minimum is the closest approach.
Condition for xi + yj to be parallel to pi + qj
x/p = y/q — one vector is a multiple of the other.
B is due north of A
The i-components of r_A and r_B are equal (and the j-component of r_B − r_A is positive).
Showing that two moving objects collide
Both components of r_B − r_A are zero at the same value of t.
Least distance between two objects without calculus
Write d² = |r_B − r_A|² as a quadratic in t and complete the square: the vertex gives the time and the least d².
How modelling a surface as smooth affects an answer
Friction is ignored, so a calculated acceleration is too large (or a required force too small) compared with reality.
Velocity (−6i + 2.5j) km h⁻¹ (i east, j north). Speed and bearing?
Speed √(36 + 6.25) = 6.5 km h⁻¹; bearing 270° + tan⁻¹(2.5/6) = 293°.
A 20 N force on a bearing of 240°. In i, j form?
20 sin 240° i + 20 cos 240° j = −10√3 i − 10 j ≈ (−17.3i − 10j) N.
Resultant of (2i + 3j) and (pi − 7j) is parallel to 3i − 2j. Find p.
(2 + p)/3 = −4/−2 = 2, so p = 4.
r = (−4 + 5t)i + (2 + 3t)j. When is it due north of the point (6, 3)?
−4 + 5t = 6 → t = 2; then r = 6i + 8j, which is 5 units due north.
Resultant force (6i − 8j) N on a 2 kg particle. Magnitude of acceleration?
a = 3i − 4j, so |a| = 5 m s⁻².
u = (−3i + 4j) m s⁻¹, constant a = (2i + 2j) m s⁻². Speed at t = 3 s?
v = 3i + 10j, speed √109 = 10.4 m s⁻¹.
d² = 20t² − 112t + 160. Least distance and when?
d² = 20(t − 2.8)² + 3.2, so least d = √3.2 = 1.79 at t = 2.8.
tan⁻¹(y/x) for a vector xi + yj with x < 0
The calculator answer points the opposite way: add 180° (sketch first).
“Due east of O” versus “moving due east”
Due east of O: j-component of r is zero. Moving due east: j-component of v is zero (and its i-component positive).
Two paths cross — is that a collision?
Not necessarily: they must be at the crossing point at the same time.
Explaining a modelling assumption
Name the object from the question and say what is ignored, e.g. “the sledge's size is ignored, so it is treated as a point”.

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