# Search results

1. ### Oscilation help

Yes. It is a function the shear modulus, G (which is related to Young's modulus by Poisson's ratio for isotropic materials) and the geometry of the cross section. KT = GJ(x) (J(x) is the polar moment of inertia of the cross section). For example, for a cylinder, the cross section is a...
2. ### Oscilation help

Consider a simple physical system composed of a mass attached to a (grounded) spring. If you were to pull the mass a distance x, the restoring force from the spring with a magnitude equal to |kx|, where k is the stiffness of the spring and has units of N/m. This comes from Hooke's law. Say...
3. ### Oscilation help

The k in your first equation is a torsional stiffness with units (Nm)/rad The k in your second equation is a linear stiffness with units N/m If you work out the units, you will see that they both work out properly. Both equations come from the equations of motion for each system (meaning...
4. ### Hollow Pipe vs Solid Rod

Nicholas, please don't double post. I also responded to your question here .
5. ### Definition of Lagrange function

You can tabulate the work done by non-conservative forces in a Lagrangian formulation by including generalized forces. L = T^* - U \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right) - \frac{\partial L}{\partial q_i} = Q_i (That's just the notation I remember: qi = ith...
6. ### Gear problem

I'd like to help you out, but I don't quite follow. Are you saying that you have a driven gear (pinion) that begins a gear train and that each successive gear in the train gives a 10x reduction in torque? With such a gear train, you will quickly run into a problem with the existence of a small...
7. ### Classical string problem

Oh, I thought the table was frictionless and the rope was held until t = 0, when it was released. As it is now stated, I'm not sure how to go about solving the problem. It seems to me that the initial static equilibrium conditions will not help you solve the dynamics, since you know nothing...
8. ### Classical string problem

There's probably a better way to do this, but I'd try this: F_{\rm net} = m\ddot{x} = \frac{\frac{L}{2}+x}{L}mg where x is the distance between the end of the rope on the table and its starting point. The net force on the rope is just the weight of the fraction of the rope that is...