Rotational Kinematics and Energy.  What is the kinetic energy?
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k e =1/2  * m *v *v\n    =1/2 *3 * (3.5*10) * (3.5*10)\n    =0.18j (ans)
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a) when it it closest to the gravitational source\nb) when the potential energy is minimum (same as a)\nc) speed = |velocity|\nd) momentum(p)= mv\ne) KE = (mv^2)/2\nf) usualy the potential energy is defined as 0 at infitity\ng) conservation of energy\nh) when it it closest to the gravitational source\ni) conservation of angular momentum\n\nThat should be enouph to figure out the answers... but think about it and draw some pictures so that you know it is right.
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Rotational mechanics is a huge field and this space wouldn't be enough to explain it. Neverthless you can see this site for the topics it contains: http://www.batesville.k12.in.us/physics/phynet/Mechanics/RotMechanics/Rotate_Mech_Intro.html\n\nIf you want to know a specific think about Rotational Mechanics then ask that. If you don't know anything about it, then I suggest you buy a book and read about it.
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http://en.wikipedia.org/wiki/Kinetic_energy
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Kinetic Energy (KE) is calculated as,\nKE = 1/2 m*v^2\nwhere m is the object's mass, and v is the object's velocity.\n\nIt is assumed that the velocity (and thus KE) is calculated with respect to one's own reference frame as there are no absolute velocities in the universe.\nThus, a moving object does contain KE.\n\nThere are many types of Potential Energy.  There is gravitational PE, Electrical PE, Chemical PE, Elastic PE, ....\nSo it is very likely for an object to posses at least one type of PE, but at the same time, it is not required (at least in theory).\nPotential Energy is measured with respect to some kind of "Ground" or base on which the object has Potential over...if the object theoretically were the base, it would have zero potential.\nI would not go so far to make the statement that every stationary object has Potential Energy, but if you looked hard enough and considered various types of PE, you might be able to find something in most cases.
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No because of Kinetic energy.\n\nLesson 1: Basic Terminology and Concepts\nKinetic Energy\nKinetic energy is the energy of motion. An object which has motion - whether it be vertical or horizontal motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the energy due to vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to motion from one location to another). To keep matters simple, we will focus upon translational kinetic energy. The amount of translational kinetic energy (from here on, the phrase kinetic energy will refer to translational kinetic energy) which an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. The following equation is used to represent the kinetic energy (KE) of an object.
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Da_vinci's answer is partly correct but it confuses force, power and energy. Energy is the integral of power over time, or more simply, if power is constant, energy is power * time. Kinetic energy can also be expressed as .5*M*V^2, M being mass and V being velocity.\nDa_vinci's example illustrates that since velocity is measured with respect to some reference, so is kinetic energy. Without a reference an accelerated object receives a velocity change but the kinetic energy change cannot be defined. Similarly, without a reference, decelerating the object requires application of a force but does not recover any energy.
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There is no buildup of kinetic energy.  You simply swing your arms to counteract the torque caused by throwing your leg forward.
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One the surface, this appears to be a fairly difficult and mathmatically intense problem, but when you look at it more closely, it really only involves a few calculations.\n\nThe underlying concept is of course conservation of energy.  The yo-yo is allowed to drop a given distance, thus it experiences a change in potential energy.  The potential energy given up is transfered into both translational kinetic energy and rotational kinetic energy.\n\nWe know that the change in Potential energy = the change in total kinetic energy.  We know that the rotational KE + translational KE = total KE.\nPE = mgh\nTranslational KE = 1/2 m*v^2\nRotational KE = 1/2 I*w^2
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Force times distance is not kinetic energy. Force times distance is work. Work can appear in different forms. If you expend 100 Newton-meters of work pushing a dog very slowly up a frictionless inclined plane, the dog will have 100 Newton-meters of potential energy when you are done, and no kinetic energy, for it will stop moving when you stop pushing.\n\nOn the other hand, if you expend 100 Newton-meters of work pushing the dog along a horizontal frictionless slippy-slide, the dog will continue sliding after you stop. It will have no potential energy and 100 Newton-meters of kinetic energy.\n\nDo you want to know how fast the dog will be going when you stop pushing? Use\n\nE = (1/2) * m *  v^2
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kinetic energy is 1/2 mv² so it depends on mass and velocity
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Kinetic Energy describes the amount of energy an object has by virtue of it hasving some non-zero mass and traveling at some speed.\nKE= 1/2 mv^2\n\nPotential Energy describes the energy that could be gained by that object if it was not at that potential.  It is not really accurate to say that potential energy is energy at rest since it is very possible for moving objects to have potential energy.\nA good example would be a rock sitting on a cliff.  That rock has some potential gravitiational energy since it is sitting some height above the ground (the bottom of the cliff).  The potential energy equals the amount of energy needed to put that rock up there in the first place and the amount of energy which could be gained if the rock falls down from the cliff.  As the rock is falling, before it hits the ground, the rock has some fraction of its potential energy converted into kinetic, in other words it has both at the same time.\nThere are many forms of potential energy, gravitational potential energy being the most commonly described.\ngravitational PE = mgh
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E1=M*g*h, M=(your mb)=32kg, - pot energy of the system at rest,\nh=2m and g=9.8m/s^2.\nE2=kinetic of M=32kg\nE3=kinetic of m=20kg\nE4=kinetic of pulley u=5kg\nE5=potential energy of m at the height h, E5=m*g*h\nThus energy conservation: E1=E2+E3+E4+E5.\nE2+E3=(M+m)*V^2/2, V to be found.\nE4=I*W^2/2, angular speed being W=V/r, r=0.1m, and I=u*r^2/2 for your pulley.\nHence 2*M*g*h=(M+m)*V^2+(u*r^2/2)*(V*r)^2+2*m*g*h,\n(M+m+u/2)*V^2=2*(M-m)*g*h, - does not depend on r.\nV=2.08m/s, don’t believe & check it! – good picture gal – a piece of art.
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Assuming the rod falls within the same plane as the track, you can use conservation of energy as suggested by the previous answer. That's because there's effectively only one degree of freedom; the horizontal position of the center of mass (center of the rod) does not move. \n\nThis can be taken as the angular orientation (theta) of the rod above the horizontal. Defining w (omega) = d theta/dt, the total kinetic energy is the sum of the rotational and vertical translational kinetic energies:\n\nT = (1/2)(ML^2/12) w^2 + (1/2)M (L^2/4)cos^2(theta) w^2\n\nIn particular for theta = 0, it's just:\n\nT = (1/2)(1/3)M L^2 w^2 \n\nAnother way of seeing this is that it's just the rotational kinetic energy about the end (rather than the center) which is the total energy when it's very close to the floor. (M/3)L^2 is the moment of inertia about one end of the rod).\n\nSetting it equal to the potential energy:\n\n(1/2)(1/3)M L^2 w^2 = M g (L/2)\n-->\nw = sqrt(3g/L)\n-->\nv = wL = sqrt(3gL) = sqrt(3*9.8*1.6)\n\nThis the tangential speed with respect to the floor (the speed w.r.t. the center of mass is half that).
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The first answer is correct.  Energy.  Energy is defined as force applied through a distance.  This specific type of energy is known as kinetic energy (energy of motion).  Energy is measured in joules, or J.  E = F x D, so if you applied 50 Newtons and the object moved 1 metre, you would have expended 50 joules.
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The kinetic energy changes proportionately to the dot product of the force and the direction of motion.  If the object is always moving at right angles to the force, there is no change in kinetic energy.\n\nThe most obvious example is a planet in a circular orbit around a star.  The star exerts a constant force on the planet, but the planet's kinetic energy never changes because the force is always perpendicular to the direction of motion.\n\nThis is a case where a non-zero net force results in no change in kinetic energy.
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For moving objects, calculate kinetic energy using the formula:\nKE = (mass x velocity2)/2 or 1/2 mv2\n\nhttp://www.gmhsscience.com/problems/kineticpotential.htm