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    (Original post by langlitz)
    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}
    Thank you for your time and intuitive questions.

    Of no relevance: What do think are the prerequisites to transition into advanced books like Arnold's / Goldstein's/ Landau and Lifshitz's Classical Mechanics from the basic ones like Morin's Introduction to Classical Mechanics? Is one forced to take a break and study some maths book? Thank you.
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    (Original post by tangotangopapa2)
    Spoiler:
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    sqrt (L/g) arc cosh (L/ (L-D))
    LOL this is on my 'warm up' problem sheet at oxford xD ( in freshers ) XD
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    (Original post by tangotangopapa2)
    Attachment 583764
    y = [email protected] - 0.5gt^2
    x = [email protected] => t = x/([email protected])

    So y = [email protected][x/([email protected])] - 0.5g[x/([email protected])]^2
    => y = [email protected] - 0.5gx^2/([email protected])^2

    At y = 0 (t > 0): 0 = [email protected] - 0.5gt
    => 0.5gt = [email protected] => t = ([email protected])/g

    So x (max) = [email protected] [([email protected])/g]
    => x (max) = (2v^2/g)[email protected]@

    Area = Int [email protected] - 0.5gx^2/([email protected])^2 dx
    = (0.5x^2)[email protected] - gx^3/[6([email protected])^2] + c
    = x^2 {[email protected] - gx/[6([email protected])^2]} + c

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

    Area = (4v^4/g^2)([email protected]@)^2 {[email protected] - (2v^[email protected]@)/[6([email protected])^2]}
    = (4v^4/g^2)([email protected]@)^2 [[email protected] - (1/3)[email protected]]
    = (4v^4/g^2)([email protected]@)^2 (1/6)([email protected]/[email protected])
    = (2/3)(v^4/g^2)([email protected])([email protected])^3

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

    Max A, dA/[email protected] = 0: 3([email protected])^2([email protected])^2 - ([email protected])^4 = 0
    => ([email protected])^2 [3([email protected])^2 - ([email protected])^2] = 0
    => [email protected] = 0, ([email protected])^2 = 3 => [email protected] = (+/-) Sqrt3

    Only [email protected] = Sqrt3 satisfies 0° < @ < 90°:
    So @ = tan^-1(Sqrt3) => @ = 60°
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    (Original post by Physics Enemy)
    y = [email protected] - 0.5gt^2
    x = [email protected] => t = x/([email protected])

    So y = [email protected][x/([email protected])] - 0.5g[x/([email protected])]^2
    => y = [email protected] - 0.5gx^2/([email protected])^2

    At y = 0 (t > 0): 0 = [email protected] - 0.5gt
    => 0.5gt = [email protected] => t = ([email protected])/g

    So x (max) = [email protected] [([email protected])/g]
    => x (max) = (2v^2/g)[email protected]@

    Area = Int [email protected] - 0.5gx^2/([email protected])^2 dx
    = (0.5x^2)[email protected] - gx^3/[6([email protected])^2] + c
    = x^2 {[email protected] - gx/[6([email protected])^2]} + c

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

    Area = (4v^4/g^2)([email protected]@)^2 {[email protected] - (2v^[email protected]@)/[6([email protected])^2]}
    = (4v^4/g^2)([email protected]@)^2 [[email protected] - (1/3)[email protected]]
    = (4v^4/g^2)([email protected]@)^2 (1/6)([email protected]/[email protected])
    = (2/3)(v^4/g^2)([email protected])([email protected])^3

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

    Max A, dA/[email protected] = 0: 3([email protected])^2([email protected])^2 - ([email protected])^4 = 0
    => ([email protected])^2 [3([email protected])^2 - ([email protected])^2] = 0
    => [email protected] = 0, ([email protected])^2 = 3 => [email protected] = (+/-) Sqrt3

    Only [email protected] = Sqrt3 satisfies 0° < @ < 90°:
    So @ = tan^-1(Sqrt3) => @ = 60°
    Thanks for tagging me in this, and I managed to get the same answer as you
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    (Original post by tangotangopapa2)
    Attachment 583736
    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
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    (Original post by Physics Enemy)
    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
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    (Original post by Integer123)
    Could you explain why it is 2Tr rather than just Tr please? Thanks
    Piece of rope either side of the pulley, each piece has an upward tension acting on the pulley.
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    (Original post by Physics Enemy)
    There's a piece of rope either side of the pulley, each piece has an upward tension acting on the pulley.
    Thanks, I thought Tr was the rod (rather than the rope) tension which confused me
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    (Original post by tangotangopapa2)
    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

    Attachment 584250
    Attachment 694422

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

    Take moments about P: L.Nsc = [email protected](L/2)
    => N = (L/2)[email protected]
    => N = [email protected](1 + [email protected])/([email protected])

    ∑ circle horiz forces: [email protected] + Fcg = [email protected]
    => F(1 + [email protected]) = [email protected]
    => F = [email protected]/(1 + [email protected])
    => F = [[email protected](1 + [email protected])/([email protected])][email protected]/(1 + [email protected])
    (1 + [email protected])/[email protected] cancels on top/bottom:
    => Fcg = ([email protected])/2

    2) i) Fsc ≤ Usc.Nsc
    => Usc ≥ Fsc/Nsc = ([email protected]/2)/[[email protected](1 + [email protected])/([email protected])]
    ([email protected])/2 cancels on top/bottom:
    => Usc ≥ 1/[(1 + [email protected])/[email protected]]
    => Usc ≥ [email protected]/(1 + [email protected])

    ii) Fsg ≤ Usg.Nsg

    ∑ stick vert forces: Nsg + [email protected] + [email protected] = pLg
    => Nsg + [email protected] + [email protected] = Rpg(1 + [email protected])/[email protected]
    => Nsg + ([email protected])^2[Rpg(1 + [email protected])/([email protected])] + [email protected]([email protected])/2 = Rpg(1 + [email protected])/[email protected]
    => Nsg = [Rpg/([email protected])] [2(1 + [email protected]) - [email protected]([email protected])^2 - (1 + [email protected])([email protected])^2]
    => Nsg = [Rpg/([email protected])] {2 + [email protected] - [email protected][1 - ([email protected])^2] - ([email protected])^2 - ([email protected])^3}
    => Nsg = [Rpg/([email protected])] [2 + [email protected] - cos(@)^2]
    => Nsg = [Rpg/([email protected])] (2 - [email protected])(1 + [email protected])

    ∑ stick horiz forces: Fsg + [email protected] = [email protected]
    => Fsg = [email protected] - [email protected]
    => Fsg = [email protected][[email protected](1 + [email protected])/([email protected])] - [email protected]([email protected])/2
    => Fsg = [Rpg/([email protected])] [[email protected]@(1 + [email protected]) - [email protected]([email protected])^2]
    => Fsg = [Rpg/([email protected])] ([email protected]@)(1 + [email protected] - [email protected])
    => Fsg = [Rpg/([email protected])] ([email protected]@)

    So [Rpg/([email protected])] ([email protected]@) ≤ Usg.[Rpg/([email protected])](2 - [email protected])(1 + [email protected])
    => ([email protected]@) ≤ Usg.(2 - [email protected])(1 + [email protected])
    => Usg ≥ ([email protected]@)/[(2 - [email protected])(1 + [email protected])]
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    (Original post by tangotangopapa2)
    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 com: m1r1= m2r2 => m1/m2 = r2/r1

    Note if m1 =/= m2, com is different (opposite end) to g = 0 point, where m1/m2 = (r1/r2)^2.
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    (Original post by Physics Enemy)
    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
    (Original post by Physics Enemy)
    Attachment 673550

    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)/[email protected]/[2cos(@/2)]
    => L = 2R[cos(@/2)]^2/[email protected]
    => L = 2R[(1 + [email protected])/2]/[email protected]
    => L = R(1 + [email protected])/[email protected]

    Take moments about P: L.Nsc = [email protected](L/2)
    => N = (L/2)[email protected]
    => N = [email protected](1 + [email protected])/([email protected])

    Sum circle's horiz forces: [email protected] + Fcg = [email protected]
    => F(1 + [email protected]) = [email protected]
    => F = [email protected]/(1 + [email protected])
    => F = [[email protected](1 + [email protected])/([email protected])][email protected]/(1 + [email protected])
    (1 + [email protected])/[email protected] cancels on top/bottom:
    => Fcg = ([email protected])/2

    2) i) Fsc ≤ Usc.Nsc
    => Usc ≥ Fsc/Nsc = ([email protected]/2)/[[email protected](1 + [email protected])/([email protected])]
    ([email protected])/2 cancels on top/bottom:
    => Usc ≥ 1/[(1 + [email protected])/[email protected]]
    => Usc ≥ [email protected]/(1 + [email protected])

    ii) Fsg ≤ Usg.Nsg

    Sum stick's vert forces: Nsg + [email protected] + [email protected] = pLg
    => Nsg + [email protected] + [email protected] = Rpg(1 + [email protected])/[email protected]
    => Nsg + ([email protected])^2[Rpg(1 + [email protected])/([email protected])] + [email protected]([email protected])/2 = Rpg(1 + [email protected])/[email protected]
    => Nsg = [Rpg/([email protected])] [2(1 + [email protected]) - [email protected]([email protected])^2 - (1 + [email protected])([email protected])^2]
    => Nsg = [Rpg/([email protected])] {2 + [email protected] - [email protected][1 - ([email protected])^2] - ([email protected])^2 - ([email protected])^3}
    => Nsg = [Rpg/([email protected])] [2 + [email protected] - cos(@)^2]
    => Nsg = [Rpg/([email protected])] (2 - [email protected])(1 + [email protected])

    Sum stick's horiz forces: Fsg + [email protected] = [email protected]
    => Fsg = [email protected] - [email protected]
    => Fsg = [email protected][[email protected](1 + [email protected])/([email protected])] - [email protected]([email protected])/2
    => Fsg = [Rpg/([email protected])] [[email protected]@(1 + [email protected]) - [email protected]([email protected])^2]
    => Fsg = [Rpg/([email protected])] ([email protected]@)(1 + [email protected] - [email protected])
    => Fsg = [Rpg/([email protected])] ([email protected]@)

    So [Rpg/([email protected])] ([email protected]@) ≤ Usg.[Rpg/([email protected])](2 - [email protected])(1 + [email protected])
    => ([email protected]@) ≤ Usg.(2 - [email protected])(1 + [email protected])
    => Usg ≥ ([email protected]@)/[(2 - [email protected])(1 + [email protected])]
    Thanks for these. I haven't learn either of these topics yet (my only mechanics knowledge comes from AS Physics - we're doing M1 and M2 in maths next year) but I'm going to take a look at both topics tomorrow and then attempt the questions
 
 
 
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