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    (Original post by zetamcfc)
    D3+ would start to get interesting actually
    Omg triggered


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    outer surface....do you mean the external one? Are the available forces adequate for such a motion? I am missing sth.
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    (Original post by drandy76)
    Omg triggered


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    C'mon you could do some proper graph theory and other stuff, it would be fun.
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    OK, I'm struggling to finish this off at the moment but my approach was as follows:

    Spoiler:
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    1. When the particle has fallen to a position where the cone radius is r, we have the horizontal velocity v_H = \frac{aU}{r} by conservation of AM.

    2. Now resolving perpendicular to the surface of the cone gives

    \displaystyle mg \sin \alpha -R = \frac{m v_H^2}{\rho} \Rightarrow R = m(g \sin \alpha - \frac{a^2U^2}{r^2 \rho})

    where \rho is the instantaneous radius of curvature of the cone along the plane defined by the normal and the direction of v_H i.e. the horizontal. But this plane is an ellipse, and the particle is at the vertex, so we need the radius of curvature of this ellipse at the vertex. That can be shown to be b^2/a where a,b are the semi-major, minor axes.

    Now I've found a in terms of r (I think) but b is giving me huge grief, so at the moment, I'm stymied with this approach. Note that it will agree with the stated result if \rho = r \sec \alpha, which I thought I'd shown last night, but I'd made a mistake.
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    come on exam boards bring on M8 if you think you're hard enough

    :pwnd:

    :horns:
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    Taking the origin at the vertex and z increasing downwards

    there are 4 equations
    1. transversly conservation of angular momentum
    2. radially (usual polar acceleration = Rcos(...)
    3. r = ztan (...) which can be differentiated and used where needed
    4. mz (double dot) = mg - Rsin (...)
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    (Original post by zetamcfc)
    C'mon you could do some proper graph theory and other stuff, it would be fun.
    Noooo, I barely tolerated all the definitions and simplex iterations last year, no more


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    Have a look, it is close enough but...
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  1. File Type: pdf not right.pdf (848.7 KB, 68 views)
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    (Original post by TeeEm)

    mz (double dot) = mg - Rsin (...)
    This is where I'm having problems. I've drawn a diagram of the situation as described in the question, with the particle on the outer surface of the inverted cone. In my diagram, mz (double dot) is not mg - Rsin (...) but rather
    -mg - Rsin (...) (i.e. both are minus and not plus and minus, if we take the downward to be positive then both terms will be positive). Can someone clear this up? I currently have the correct answer except the two terms have the same sign and not different signs.
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    (Original post by A Slice of Pi)
    This is where I'm having problems. I've drawn a diagram of the situation as described in the question, with the particle on the outer surface of the inverted cone. In my diagram, mz (double dot) is not mg - Rsin (...) but rather
    -mg - Rsin (...) (i.e. both are minus and not plus and minus, if we take the downward to be positive then both terms will be positive). Can someone clear this up? I currently have the correct answer except the two terms have the same sign and not different signs.
    the cone is the other way round
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    (Original post by TeeEm)
    the cone is the other way round
    That would make more sense, but in the question wasn't it called an 'inverted' cone?
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    (Original post by A Slice of Pi)
    That would make more sense, but in the question wasn't it called an 'inverted' cone?
    I am teaching at present to have a discussion but the cone in that orientation is inverted
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    (Original post by TeeEm)
    I am teaching at present to have a discussion but the cone in that orientation is inverted
    Well, I think it's definitely debatable. I'd maybe include a diagram in questions like this
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  2. File Type: pdf solution2.pdf (191.2 KB, 68 views)
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    (Original post by depymak)
    Have a look, it is close enough but...
    Your method is very good . I think its actually the angle 'alpha' in the diagram that is incorrectly labelled, which is a shame because the rest of it is perfectly correct
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    My solution
    Name:  Motion on a cone.jpg
Views: 65
Size:  37.8 KB
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    thanks, I misinterpreted the term "semi-vertical angle " !
    (Original post by A Slice of Pi)
    Your method is very good . I think its actually the angle 'alpha' in the diagram that is incorrectly labelled, which is a shame because the rest of it is perfectly correct
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    My solution:

    We locate the particle on the cone of semi-vertical angle \alpha via cylindrical polar coords r, \theta, z, with z increasing downwards. The initial angular momentum of the particle about the z-axis is J =maU and the motion of the particle is subject to the constraint:

    f(r,z)=r-z\tan \alpha = 0

    Its KE, T, and PE, V, are given by:

    T = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2)
    V = -mgz

    taking V=0 to be at the vertex of the cone, and V decreasing downwards.

    The Euler-Lagrange equations for a particle with generalised coords q_i subject to a holonomic constraint f(q_1, q_2, \cdots, q_n, t)=0 are:

    \displaystyle \frac{d}{dt}\big(\frac{\partial L}{\partial \dot{q_i}}\big)-\frac{\partial L}{\partial q_i} -\lambda \frac{\partial f}{\partial q_i} =0

    where L=T-V and where we have q_i \in {r,\theta,z}. Also, the force of constraint conjugate to q_i is:

    Q_{q_i} = \lambda \frac{\partial f}{\partial q_i}

    and these forces act in the direction of increasing q_i

    So for this problem, the E-L equations give (details omitted):

    (1) m\ddot{r} -mr\dot{\theta}^2-\lambda =0

    (2) \frac{d}{dt}(mr^2\dot{\theta}) = 0 \Rightarrow mr^2\dot{\theta} = J where J is the conserved angular momentum.

    (3) m\ddot{z} -mg+\lambda \tan \alpha =0

    Now using (2) to eliminate \dot{\theta} from (1) gives:

    m\ddot{r} - \frac{J^2}{mr^3} -\lambda =0 \Rightarrow \lambda = m\ddot{r} - \frac{J^2}{mr^3} (4)

    and using the equation of constraint to eliminate z from (3) gives:

    m\ddot{r} \tan \alpha -mg+\lambda \tan \alpha =0 \Rightarrow m\ddot{r} = \tan \alpha (mg -\lambda \tan \alpha) (5)

    So (4) and (5) give us, after some algebra:

    \lambda = \cos^2 \alpha (mg \tan \alpha - \frac{J^2}{mr^3}) =  m\cos^2 \alpha (g \tan \alpha - \frac{a^2U^2}{r^3})

    So the constraint force conjugate to r is:

    Q_r = \lambda \frac{\partial f}{\partial r} = \lambda

    But this is the constraint force in the direction of increasing r, and is thus the horizontal component of the normal reaction R so we have:

    Q_r = R \cos\alpha \Rightarrow R = \frac{Q_r}{\cos \alpha} =  m\cos \alpha (g \tan \alpha - \frac{a^2U^2}{r^3})

    as required.

    Note that my other approach would have been much quicker if it wasn't for the fact that I couldn't find the radius of curvature of the ellipse in question. So while the workings above represent another triumph for M. Lagrange, I suspect that my other method was more Newtonian in character, and probably could have been completed easily by Mr Newton, who would have known all about ellipses.
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    nice!
    (Original post by TeeEm)
    My solution
    Name:  Motion on a cone.jpg
Views: 65
Size:  37.8 KB
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    (Original post by atsruser)
    ....
    Lagrangian considerations make it at least M9 ....
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    (Original post by TeeEm)
    Lagrangian considerations make it at least M9 ....
    Well, I couldn't finish my quick 'n dirty method, and other people had done it the obvious way, so I had little choice but to bring out the big guns. Maybe hides the physical insight though, compared to getting down and dirty with the forces.
 
 
 
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