Title: InkMaths / Further Mechanics 2: circles, balance and oscillation
On every note of this deck, unless a note says otherwise:
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Front: DEFINE
Define: completing a vertical circle.
Back: On a string the tension must stay non-negative, so v² ≥ gr at the top and u² ≥ 5gr at the bottom. On a rod, which can push, the particle need only arrive at the top with v > 0.

Front: STANDARD RESULT — LEARN IT
Centre of mass by integration:
lamina: mean x = ?, mean y = ?
Back: ∫ xy dx∫ y dx, ∫ 12y^2 dx∫ y dx

Front: STANDARD RESULT — LEARN IT
Banked track:
no friction needed when tan θ = ?
Back: v^2rg: the mass cancels

Front: DEFINE
Define: lamina against framework.
Back: A uniform lamina weights each piece by its area at its own centroid, with a negative area for anything removed. A framework weights each rod by its length at its own midpoint, so the two differ for the same shape.

Front: STANDARD RESULT — LEARN IT
Centre of mass:
discrete: mean x = ?
Back: Σ m_ix_iΣ m_i: a mass-weighted average

Front: DEFINE
Define: choosing the acceleration.
Back: Use v dv/dx when the force depends on position, dv/dt when it depends on time. The constant-acceleration equations are not available at all, whatever average is taken.

Front: CHECK YOURSELF
A lamina is suspended freely from a corner. Where does its centre of mass end up?
Back: Vertically below the point of suspension, because otherwise the weight would have a moment about the pivot and the lamina would turn.

Front: STANDARD RESULT — LEARN IT
Speed in simple harmonic motion:
v^2 = ?
Back: ω^2(a^2 - x^2): fastest at the centre

Front: STANDARD RESULT — LEARN IT
Spring oscillation:
ω^2 = ?
Back: λml: centred on the equilibrium position

Front: STANDARD RESULT — LEARN IT
Topple or slide:
topples when tan θ = ?, slides when tan θ = ?
Back: ah, μ: the smaller angle happens first

Front: STANDARD RESULT — LEARN IT
Simple harmonic motion:
condition? T = ?
Back: acceleration = -ω^2x, 2πω whatever the amplitude

Front: STANDARD RESULT — LEARN IT
Circular motion:
v = ?, a = ?
Back: rω, rω^2 = v^2r towards the centre

Front: SPOT THE ERROR
Say what is wrong with this claim:
A particle going round a circle at constant speed has no acceleration.
Back: Velocity is a vector and its direction is changing continually, so there is an acceleration of rω² directed at the centre. Some real force has to supply it.