Hard Mechanics Discussion Thread

I think so I don't think you'd get all the marks for what you wrote in b) though.

p.s. to write vectors in latex you do \vec{b} for example for a vector $\vec{b}$

y = vtsin@ - 0.5gt^2
x = vtcos@ => t = x/(vcos@)

So y = vsin@[x/(vcos@)] - 0.5g[x/(vcos@)]^2
=> y = xtan@ - 0.5gx^2/(vcos@)^2

At y = 0 (t > 0): 0 = vsin@ - 0.5gt
=> 0.5gt = vsin@ => t = (2vsin@)/g

So x (max) = vcos@ [(2vsin@)/g]
=> x (max) = (2v^2/g)sin@cos@

Area = Int xtan@ - 0.5gx^2/(vcos@)^2 dx
= (0.5x^2)tan@ - gx^3/[6(vcos@)^2] + c
= x^2 {0.5tan@ - gx/[6(vcos@)^2]} + c

Upper limit = x (max), Lower limit = 0:

Area = (4v^4/g^2)(sin@cos@)^2 {0.5tan@ - (2v^2sin@cos@)/[6(vcos@)^2]}
= (4v^4/g^2)(sin@cos@)^2 [0.5tan@ - (1/3)tan@]
= (4v^4/g^2)(sin@cos@)^2 (1/6)(sin@/cos@)
= (2/3)(v^4/g^2)(cos@)(sin@)^3

dA/d@ = (2/3)(v^4/g^2) [[email protected]@(sin@)^2 - sin@(sin@)^3]
= (2/3)(v^4/g^2) [3(cos@)^2(sin@)^2 - (sin@)^4]

Max A, dA/d@ = 0: 3(cos@)^2(sin@)^2 - (sin@)^4 = 0
=> (sin@)^2 [3(cos@)^2 - (sin@)^2] = 0
=> sin@ = 0, (tan@)^2 = 3 => tan@ = (+/-) Sqrt3

Only tan@ = Sqrt3 satisfies 0° < @ < 90°:
So @ = tan^-1(Sqrt3) => @ = 60°

We want N, Tp, Tr in terms of M, m, u, g, a:

Forces on the girl: (Pull - Tr) + N - mg = ma
Inextensible, massless rope => Pull = Tr
So N - mg = ma => N = m(g + a)

On the platform: Tp - Mg - mg = Ma
=> Tp = aM + g(M + m)

On the pulley: 2Tr - ug - Tp = ua
=> Tp = 2Tr - u(g + a)

So 2Tr - u(g + a) = aM + g(M + m)
=> 2Tr = aM + u(g + a) + g(M + m)
=> 2Tr = a(M + u) + g(M + m + u)
=> Tr = [a(M + u) + g(M + m + u)]/2

Could you explain why it is 2Tr rather than just Tr please? Thanks

There's a piece of rope either side of the pulley, each piece has an upward tension acting on the pulley.

A stick of mass density per unit length ρ rests on a circle of radius R.The stick makes an angle θ with the horizontal and is tangent to the circle at its upper end. Friction exists at all points of contact and it is large enough to keep the system at rest.

1) Find the friction force between the ground and the circle.
2) Find the co-efficients of friction between: i) stick/circle and ii) stick/ground

2 components of a binary star move in circles, orbital radii r₁ and r₂. Ratio of their masses? Note the stars orbit combined com.

At the c of m: m1r1= m2r2 => m1/m2 = r2/r1

Note for m1 =/= m2 the c of m is different to the g = 0 point, where m1/m2 = (r1/r2)^2

1) Nsc = Normal on stick by circle, Fcg = Friction on circle by ground, etc
Circle is in rotational equilibrium, so net torque = 0, so Fcs = Fcg ≡ F
Equilibrium where stick/circle meet: Nsc = Ncs ≡ N and Fcs = Fsc ≡ F

Note the diagram's kite split into 2 congruent triangles:
tan(@/2) = R/L => L = R/tan(@/2) = Rcos(@/2)/sin(@/2)
=> L = Rcos(@/2)/sin@/[2cos(@/2)]
=> L = 2R[cos(@/2)]^2/sin@
=> L = 2R[(1 + cos@)/2]/sin@
=> L = R(1 + cos@)/sin@

Take moments about P: L.Nsc = pLgcos@(L/2)
=> N = (L/2)pgcos@
=> N = Rpgcos@(1 + cos@)/(2sin@)

Sum circle's horiz forces: Fcs.cos@ + Fcg = Ncs.sin@
=> F(1 + cos@) = Nsin@
=> F = Nsin@/(1 + cos@)
=> F = [Rpgcos@(1 + cos@)/(2sin@)]sin@/(1 + cos@)
(1 + cos@)/sin@ cancels on top/bottom:
=> Fcg = (Rpgcos@)/2

2) i) Fsc ≤ Usc.Nsc
=> Usc ≥ Fsc/Nsc = (Rpgcos@/2)/[Rpgcos@(1 + cos@)/(2sin@)]
(Rpgcos@)/2 cancels on top/bottom:
=> Usc ≥ 1/[(1 + cos@)/sin@]
=> Usc ≥ sin@/(1 + cos@)

ii) Fsg ≤ Usg.Nsg

Sum stick's vert forces: Nsg + Nsc.cos@ + Fsc.sin@ = pLg
=> Nsg + Ncos@ + Fsin@ = Rpg(1 + cos@)/sin@
=> Nsg + (cos@)^2[Rpg(1 + cos@)/(2sin@)] + sin@(Rpgcos@)/2 = Rpg(1 + cos@)/sin@
=> Nsg = [Rpg/(2sin@)] [2(1 + cos@) - cos@(sin@)^2 - (1 + cos@)(cos@)^2]
=> Nsg = [Rpg/(2sin@)] {2 + 2cos@ - cos@[1 - (cos@)^2] - (cos@)^2 - (cos@)^3}
=> Nsg = [Rpg/(2sin@)] [2 + cos@ - cos(@)^2]
=> Nsg = [Rpg/(2sin@)] (2 - cos@)(1 + cos@)

Sum stick's horiz forces: Fsg + Fsc.cos@ = Nsc.sin@
=> Fsg = Nsin@ - Fcos@
=> Fsg = sin@[Rpgcos@(1 + cos@)/(2sin@)] - cos@(Rpgcos@)/2
=> Fsg = [Rpg/(2sin@)] [sin@cos@(1 + cos@) - sin@(cos@)^2]
=> Fsg = [Rpg/(2sin@)] (sin@cos@)(1 + cos@ - cos@)
=> Fsg = [Rpg/(2sin@)] (sin@cos@)

So [Rpg/(2sin@)] (sin@cos@) ≤ Usg.[Rpg/(2sin@)](2 - cos@)(1 + cos@)
=> (sin@cos@) ≤ Usg.(2 - cos@)(1 + cos@)
=> Usg ≥ (sin@cos@)/[(2 - cos@)(1 + cos@)]