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: RECALL
State the condition for 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 = ?\(\text{lamina, mean x} = ?, \text{mean y} = ?\)lamina, mean x = ?, mean y = ?
Back: ∫ xy dx∫ y dx, ∫ 12y^2 dx∫ y dx\(\frac{\text{∫} xy \, dx}{\text{∫} y \, dx}, \frac{\text{∫} \frac{1}{2}y^{2} \, dx}{\text{∫} y \, dx}\)((∫ xy dx)/(∫ y dx)), ((∫ (1/2)y^2 dx)/(∫ y dx))

Front: STANDARD RESULT · LEARN IT
Banked track:
no friction needed when tan θ = ?\(\text{no friction needed when } \tan θ = ?\)no friction needed when tan θ = ?
Back: v^2rg. The mass cancels\(\frac{v^{2}}{rg}. \text{ The mass cancels}\)(v^2/rg). The mass cancels

Front: RECALL
Explain how a lamina and a framework differ.
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 distribution, mean x = ?\(\text{discrete distribution, mean x} = ?\)discrete distribution, mean x = ?
Back: Σ m_ix_iΣ m_i, a mass-weighted average\(\frac{Σ m_{i}x_{i}}{Σ m_{i}}, \text{ a mass-weighted average}\)((Σ m_ix_i)/(Σ m_i)), a mass-weighted average

Front: RECALL
State how to choose the form of 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 = ?\(v^{2} = ?\)v^2 = ?
Back: ω^2(a^2 - x^2). Fastest at the centre, at rest at the ends\(ω^{2}(a^{2} - x^{2}). \text{ Fastest at the centre, at rest at the ends}\)ω^2(a^2 - x^2). Fastest at the centre, at rest at the ends

Front: STANDARD RESULT · LEARN IT
Spring oscillation:
ω^2 = ?\(ω^{2} = ?\)ω^2 = ?
Back: λml, measured from the equilibrium position\(\frac{λ}{ml}, \text{ measured from the equilibrium position}\)(λ/ml), measured from the equilibrium position

Front: STANDARD RESULT · LEARN IT
Block of width 2a and height 2h on a slope:
topples when tan θ = ?, slides when tan θ = ?\(\text{topples when } \tan θ = ?, \text{slides when } \tan θ = ?\)topples when tan θ = ?, slides when tan θ = ?
Back: ah, μ. Whichever angle is smaller happens first\(\frac{a}{h}, μ. \text{ Whichever angle is smaller happens first}\)(a/h), μ. Whichever angle is smaller happens first

Front: STANDARD RESULT · LEARN IT
Simple harmonic motion:
condition? T = ?\(\text{condition?} T = ?\)condition? T = ?
Back: acceleration = -ω^2x, 2πω whatever the amplitude\(\text{acceleration} = -ω^{2}x, \frac{2π}{ω} \text{ whatever the amplitude}\)acceleration = -ω^2x, (2π/ω) whatever the amplitude

Front: STANDARD RESULT · LEARN IT
Circular motion:
v = ?, a = ?\(v = ?, a = ?\)v = ?, a = ?
Back: rω, rω^2 = v^2r towards the centre\(rω, rω^{2} = \frac{v^{2}}{r} \text{ towards the centre}\)rω, rω^2 = (v^2/r) 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.