![]() ![]() Integrating curvatures over beam length, the deflection, at some point along x-axis, should also be reversely proportional to I. Therefore, it can be seen from the former equation, that when a certain bending moment M is applied to a beam cross-section, the developed curvature is reversely proportional to the moment of inertia I. ![]() Beam curvature κ describes the extent of flexure in the beam and can be expressed in terms of beam deflection w(x) along longitudinal beam axis x, as: \kappa = \frac. Where E is the Young's modulus, a property of the material, and κ the curvature of the beam due to the applied load. The bending moment M applied to a cross-section is related with its moment of inertia with the following equation: The moment of inertia (second moment or area) is used in beam theory to describe the rigidity of a beam against flexure (see beam bending theory). The term second moment of area seems more accurate in this regard. This is different from the definition usually given in Engineering disciplines (also in this page) as a property of the area of a shape, commonly a cross-section, about the axis. Understanding physics deeply is an essential aspect of this pursuit. I am a theoretical physicist (M.Sc.) and my work revolves around becoming the next Albert Einstein. It is related with the mass distribution of an object (or multiple objects) about an axis. Derivation of the moment of inertia of a homogeneous hollow cylinder and a solid cylinder rotating around its axis of symmetry. In Physics the term moment of inertia has a different meaning. If you have more friction then you will have a much more complicated problem.The dimensions of moment of inertia (second moment of area) are ^4. If the amount of friction is only enough to cause the cylinders to roll without slipping then all I said above is true but if you have less friction then the cylinders will just slid down the ramp and have the same speed. Mass and radius of the cylinders should not matter. ![]() This means either increasing the length of the ramp or increasing the angle. To make this effect more noticeable is to increase the height that the cans are released from. The effect will be much less noticeable as it is will be harder to create rotation among the gas to the point if you consider the gas to be an ideal gas then there will be almost no difference. If it is a gas than it gets more complicated. This really depends on a couple different things, If the deodorant is a gas or if it is solid, what you mean by noticeable and what type of friction there is.įor the first part if you are comparing a solid deodorant than the solid deodorant can will always be sqrt(3)/2 faster than the empty can (a simple conservation of energy problem). The shallower the ramp is, the more the can has to rotate for a given drop in height, and so the more clear the difference is between a can whose motion is dominated by angular momentum (empty can) versus another whose motion is dominated by linear momentum (full can). Had the liquid been much thicker, the result would be much different, but in this case, the difference in speed between the two bottles was obvious.įor the latter part of your question, the inclined plane is what forces the can to rotate - if you just dropped them, moments of inertia wouldn't factor in at all, and the full and empty cans would drop at very close to the same rate. Also, friction between the water and the bottle wasn't enough to make much difference. However, being much more massive than the bottle, the water contributed much more to the acceleration than did the bottle itself. What was also clear is that for a low-viscosity fluid like water, there wasn't enough friction for the water to "spin up" within the bottle - it all stayed on the bottom half of the bottle throughout the journey, so the water added nothing to the moment of inertia about the bottle's axis. The empty bottle behaved as you would expect an empty cylinder to, rolling slower than the full bottle. I just did an experiment, rolling two identical transparent bottles, one empty, the other about half full, down about a 15-degree ramp. The assumption I'm making is that the fluid has a viscosity close to that of water, since the deodorant is mostly water, and by simply shaking a can of it we can tell it's not terribly viscous - it shakes like water in a can. Here we have the can, which is rolling, and the fluid, which is basically just falling. That is, a solid cylinder will roll down the ramp faster than a hollow steel cylinder of the same diameter (assuming it is rolling smoothly rather than tumbling end-over-end), because moment of inertia depends on the distribution of mass, with mass further from the axis of rotation contributing more to moment of inertia than mass closer to the axis.īut that's only part of what's in play here. As you say, "we know that hollow cylinders are slower than solid cylinders when rolled down an inclined plane". ![]()
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