Recent content by simbil

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    Separating angular and linear momentum

    Thanks Doc Al, appreciate your guidance. horizontal force = 0.75gsinθcosθ - sinθ(2g - 2gcosθ)
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    Separating angular and linear momentum

    Do I need to add more of my working to make it clearer to follow?
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    Separating angular and linear momentum

    Finally got around to this again and I have worked out what may be a solution. So, following your advice to find the velocity of the centre of mass as a function of angle: We know that the energy E is constant such that the gravitic potential and kinetic are constant so E = mgh - 0.5mv^2...
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    Separating angular and linear momentum

    Doc Al, Applying conservation of energy, I suppose we can say that the energy of the system at the start is its gravitic potential energy. At any point during the fall, the total energy will still be the same magnitude, but will be made up of a component of reduced gravitic potential energy...
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    Separating angular and linear momentum

    I think I have some fundamental confusion here. I understand that net external forces are needed to produce a change in momentum (Newton's second). I run into difficulty with cause and effect. To me it seems that the falling rod's weight causes GRF so to count the effect of gravity and the...
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    Separating angular and linear momentum

    My presumption "I presumed the horizontal component of the tangential force (the component of the rod's weight perpendicular to the rod) would be equal to the horizontal component of the GRF.", was based on Newton's second law. The force that creates the change in horizontal momentum must be the...
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    Separating angular and linear momentum

    Yes that's what I meant. I presumed the horizontal component of the tangential force (the component of the rod's weight perpendicular to the rod) would be equal to the horizontal component of the GRF. OK, so for a 2m long rod weighing 1kg, torque would be r * F = 1m * 1kg * g * sinθ = g...
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    Separating angular and linear momentum

    Doc Al, Yes, sorry you did mention that the magnitude of GRF is not equal to the rod's weight. To be precise maybe I should have said horizontal GRF is proportional to m.g.sin(Θ) rather than equal to it? I've seen hGRF = m.g.sin(Θ) mentioned before and looking at its derivation, they start...
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    Separating angular and linear momentum

    Yes, I mean the horizontal component of GRF, excuse my wandering language :) For the scenario of a simple leaning falling rod: Gravity acts on the COM and it's force can be expressed as m.g acting vertically downwards. There is a GRF acting at some angle (Θ) in the rod COM's quadrant...
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    Separating angular and linear momentum

    Thanks Doc Al. I appreciate that gravity acts vertically. My confusion and questions arise from the fact that when a rod falls over, it's COM is lowered and also moved horizontally so gravity is instrumental in a horizontal movement - when it acts on a leaning rod it creates a non-vertical...
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    Separating angular and linear momentum

    Thanks Doc Al, that makes sense and I can see why I cannot assume that the GRF will act along the rod. Onto my last point as I posted above. To abstract the runner into an easier physical model, we can consider a leaning rod (not vertical or horizontal) on a trolley. If the trolley is...
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    Separating angular and linear momentum

    OK, as no answer is forthcoming, I guess what I asked has no simple answer. To wrap this question up, I will go on to the main issue. I am discussing running technique with a number of people and particularly the role of gravity in running. Gravity is neutral over an entire running cycle...
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    Separating angular and linear momentum

    Doc Al, OK, seems like I am on the wrong track then. So if take the example of a leaning rod, how do we calculate the GRF's magnitude and direction assuming that the ground does not yield and that friction is sufficient to stop the rod slipping?
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    Separating angular and linear momentum

    Doc Al, Yes, GRF is ground reaction force, the force exerted by the ground at the contact point with an object. I presumed that the GRF would be equal in magnitude to the rod's weight but with its direction along the rod. When the rod is vertical, GRF and weight cancel and so there is no net...
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    Separating angular and linear momentum

    Doc Al, I did wonder about that as I typed it. Is it actually the case that the GRF acts along the rod with magnitude equal to the weight of the rod?
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