im looking for a real aplication of moments of inertia (Phisics)?
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Watch skating at the Olympics tonight.  When the skater brings her arms in closer to her body, she decreases her moment of inertia.  But by Conservation of Energy, her kinetic energy must remain constant.  Her kinetic energy is given by the formula\n\nK = 1/2 I w^2  , where I is the moment of inertia and w is her (angular) rotational velocity.  If I decreases, then w must increase, i.e. she spins faster as she brings her arms in.\n\nAnother example using a more conventional rigid body: some mechanical systems, and even some experimental cars, store energy in a flywheel.  The greater the moment of inertia of the flywheel, the greater the amount of energy that is stored for a given rotational speed.
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When looking at analogous equations between linear motion and rotational motion, the analogue of mass is the moment of inertia in a rotating system.\n\nFor a single point of mass, the moment of inertia is equal to the mass of the object times the square of the distance to the rotational axis.\n\nFor an arbitrary body, you need to integrate [the product of the density and (the distance to the rotational axis squared) ] over the entire volume of the body.  Since it's hard to write math notation here, I suggest you look at the formula on the source pages (it's the first formula on the scienceworld source).\n\nFor certain common shapes, formulas are given on the pages listed by the other answerers, as well as, for example, here: http://en.wikipedia.org/wiki/List_of_moments_of_inertia
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Try,\n\nhttp://www.personal.psu.edu/faculty/d/x/dxh40/MBLemma.pdf
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phi1 = atan(60/120) = atan(0.5) = 26.57 degr\nphi2 = atan(120/60) = atan(2) = 63,43 degr
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these are data from end fixing of the structure you should introduce to solve the system of equations, otherwise you would get infinite solutions or (rigide body motion)-Good Luck
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Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!
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http://eta.physics.uoguelph.ca/tutorials/torque/Q.torque.inertia.html\n\ngives 2I=mr^2  so I = 163*0.306*0.306/2\n\nhttp://buphy.bu.edu/~duffy/py105/notes/AngularMo.html\n\ngives 2 * kinetic energy = Iw^2\n\nso KE at 1500rpm = I*4/2*pi^2*(1500/60)^2 joules - everything is frictionless so this is the work done in accelerating the flywheel to 1500 rpm\n\nyou can work out the KE at 1181 rpm and find the difference that was dissipated as heat.\n\nbest of luck - Mike
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The inertia is said to be the tendency of a body to be in its state of rest or of uniform motion.\n  \nTo be frank, I don’t understand the meaning of the term “tendency” of lifeless bodies. \n\nBut by ‘Inertia’, I do mean the following.\n\nWhen a body is at rest its inertia is measured by its mass and its momentum is zero.\n\nWhen it is in uniform motion its inertia is measured by its momentum (mass x velocity)\n\nWhen it is in uniform circular motion its inertia is measured by the product of mass and the perpendicular distance from the point about which there is circular motion. Or we can say it is the moment of mass about that point.\n\nMoment measures the rotating effect of any quantity; it is the product of that quantity and the perpendicular distance of that quantity from that point. Greater is the product greater is the rotating effect of that quantity.\n\n Like linear momentum in uniform motion, angular momentum is in circular motion. Angular momentum is the moment of linear momentum.\n\nIn linear motion, to change the momentum of a body a force is necessary; whereas in circular motion to change the angular momentum a torque (moment of force) is necessary. \n\nIf m is the mass, v is the velocity and r is the perpendicular distance, then the angular momentum is mvr = mrv. But v = r w, w is the angular velocity.\n\nTherefore, angular momentum = mrrw. But we know mr is the inertia of the body in circular motion. The quantity mrr is therefore the moment of Inertia = I.\n\nTherefore, angular momentum is I w. It is similar to the linear momentum which is mv.\n\nLinear momentum is the product of  inertia at rest and velocity.\n\nAngular momentum is the product of  moment of inertia ( in circualar motion) and the moment of  linear velocity.\n\nThe moment of inertia is the moment of moment of mass about any point.
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gravity and inertia\n\nread more below. It is really fascinating.
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Try here:\n\nhttp://en.wikipedia.org/wiki/Accelerometer
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http://www.machinevibration.com
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The moment of inertia of the turntable plus the child at radias R is I + m*R^2. Setting up the equation of conservation of momentum\nI*(1 rev/T) = (I + m*R^2)*w Solve for w.
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http://online.cctt.org/physicslab/content/EssayAnswers/search.asp\n\nor \n\napcentral.collegeboard.com
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Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
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This equation only works for small angles theta < 10 degrees\n\nPeriod = 2*pi * sqrt (length / g)\n\nThe period in an undamped simple pendulum does not depend on mass.\n\nThere is a more in depth formula for larger angles on hyperphysics.
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The moment of inertia of a sphere is:\n\n(2/5)* M * R^2\n\nFor a hollow sphere (spherical shell) you can subtract the hollow part from the sphere part, but since the M (mass) term will change based on diameter it is easier to substitute:\n\nM = p * 4/3 * PI * R^3     (p = density)\n\nSo.. in the end you get  I = 0.5333 p ( R2^5 - R1^5 ) \n\nWhere R2 = Outer Radius, R1 = Inner Radius
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It is depended on the moment and the surface moments if inertia of cross section of cantilever.\n\nMoment is force time distance : M=r*f\n\nand I is used to show surface moments if inertia.\n\nthe more the moment more is the deflection.\nthe more the I less is the deflection.