kinetic energy with a rolling cylinder?
817719
The translational energy is equal to 0.5 * mass * (velocity)^2, or\n\n0.5 * 12 * 12^2 = 864 J\n\nThe rotational kinetic energy is equal to 0.5 * (moment of inertia) * (angular velocity)^2.\n\nThe moment of inertia for a cylinder is 0.5 * (mass)*(radius)^2\n\nThe angular velocity is the (velocity)/(radius).\n\nso substituting in gives:\n\n0.5 (0.5 * mass * radius^2) * (velocity/radius)^2\n\nthe radius^2 factor will cancel leaving\n\n0.25 * mass * velocity^2 or\n\n0.25 * 12 * 12^2 = 432 J\n\nTotal energy is the sum = 432 + 864 = 1296 J.
253742
Please take the time to\nvote for the best answer\n\nThank You!!\n\nRecall that an object of a certain mass moving with particular speed will have an associated kinetic energy  mass x speed2. An object spinning about an axis will also have associated with it a kinetic energy, composed of the kinetic energies of each individual part of the object. These individual contributions may be summed up to give an expression for the total kinetic energy of the spinning object: \n\nLook this up for more
1128176
work = change of kinetic enegy\nW = 1/2 m (vf² - vi²)\nW = 0.5 * (0.250 kg) * (16² - 20²) m²/s²\nW = -18 J
223530
How about the adding of Gravitational Potential Energy (GPE) and Kinetic Energy (KE) as an object falls in the air to demonstrate the preservation of energy?\n\nBefore it falls all energy is GPE, as it falls GPE is converted to KE until just before it hits the floor then it is all KE.
149886
Hope this helps,  first remember that energy must be conserved so E initial will equal E final so you use the equations for potential and kinetic energy.  mgh for potential and 1/2mv^2 for kinetic.  You know the initial velocity, gravity and the height of the ramp so you simply break it into components and get the answer of 3.15 m/s.  HOpe this helps
573522
kinetic energy is 1/2 mv² so it depends on mass and velocity
1018182
If by rolling road you mean a setup similar to a dynamometer, I assume you want to couple a load to the rollers. One way to do this would be to force a rubber-tired wheel on a spring-loaded hinged axle against the rollers, and mount some sort of drag device like a spring-loaded hinged metal bracket so it presses on the tire.
247664
It's at the apex of the circle.\n\nThe total energy of the system is the sum of the kinetic energy, 1/2 mv^2, and the potential energy, mgh (where h is the height above the circle's lowest point).  At the bottom of the circle, all the energy is kinetic and the speed is highest; at the top, some of this energy has been stored as potential energy, according to this:\n\n1/2 m (v_bottom)^2 + 2mgr = 1/2 m (v_top)^2 \n\nThe actual value of the velocity is undetermined, as v_bottom can be as fast as you like.\n\n>>>EDIT:\n\nPoster above is completely right.  Whoops :)  That'll teach me to do this on scraps of paper at work...
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.
906577
Hi!\n(A) \nBy definition di=dm * X^2 , where di is the amount added to I when the mass dm is added to the system. I~Inertia . In this case for the particle X=d.\n\nEach mass m=0.85 kg add to the system (delta I)1 = m * d^2.\n\nEach rod has an Inertia of (delta I)2 = (1/3) M * d^2 . This is because for a rod you have to calculate:\n             S { dm * x^2 = (M/d)*dx * x^2 } ==> (M/d) * (1/3)(X^3) with X=d. => (1/3)M * d^2  \n\nso the total inertia of the system is I= 2*[ m*d^2+ (1/3) M*d^2] because we have 2 rods and tho particles.:\n    I= 7.84 * 10^(-3) kg / m^2.    Where ^ is used for power!\n\n(B) kinetic energy= (1/2) Inertia  * (angularspeed)^2 =3.528 /10^4\n\nIf you want to know more try the link below!
889727
To find the tangential velocity, we can use the formula\n\nV= r w\n\nwhere r is the radius and w is the angular velocity. Since we know r all we need to do is convert w to rad/sec to find V.\n\nw = (50,000 rev/min)(2 Pi rad/rev)(1 min)/(60 sec)\n= 5235.99 rad/sec\n\nNow we have everything needed to find V.\n\nV = (.25 m)(5235.99 rad/sec)\n= 1309 m/s\n\nSo the tangential velocity is 1.3 km/s\n\nThe energy is stored in the kinetic energy of its rotation. The formula for this energy is\n\nK = (1/2) I w^2\n\nwhere I is the moment of inertia and w is the angular velocity. From the description of this flywheel, it sound like it is a loop rotating about a central axis. The rotational inertia of this shape is\n\nI = M R^2\n= (50 kg) (0.25 m)^2\n= 3.125 kg m^2\n\nUsing our previous calculation we know w in rad/sec.\n\nSo now we have all the information we need.\n\nK = (1/2)(3.125 kg m^2)(5235.99 rad/sec)^2\n= 4.3 x 10^7 J = 43 MJ\n\nThe amount of work that could be done by the flywheel assuming that all energy could be converted. The amount of work is the difference in the kinetic energy. Assuming all of the energy could be converted, this would be the total amount of work.\n\nW = Kf - Ki\n\nwhere Kf is the final kinetic energy and Ki is the initial kinetic energy.\n\nKf = 0\nKi = 43MJ\n\nW = (0) - (43MJ)\n= -43MJ\n\nIt should be noted that this is the amount of work required to stop the rotation of the flywheel, so the absolute value of this is the total amount of work that the flywheel can perform.\n\nW = |-43MJ|\n= 43MJ\n\nIf we can not get any work below 10000 rev/min, we can use the same formula. We just need to find out K for that speed. Recall that\n\nK = (1/2) I w^2\n\nI is still the same, so all we need to do now is change w to rad/sec.\n\nw = (10000 rev/min)(2 Pi rad/sec)(1 min)/(60 sec)\n= 1047.2 rad/sec\n\nSo now we have the information to find Kf\n\nKf = (1/2)(3.125 kg m^2)(1047.2 rad/sec)^2\n= 1.7 x 10^6 J = 1.7 MJ\n\nSo now we can find W\n\nW = |Kf - Ki|\n= | (1.7MJ) - (43MJ) |\n= 41MJ\n\nIf we are only able to get 75% of this work, then we must multiply this number by 75%\n\nWact. = (.75)(41MJ)\n= 31MJ\n\nwhere Wact is the actual work. The rest of the energy goes to friction of the wheel spinning on it's axis, probably on bearings or whatnot, to friction with the air around it, imperfections in the disc also cause it to be off balance and wobble, loosing energy. It goes lots of places. But friction is probably the answer you're looking for.\n\nI hope this is right, it's been a while since my physics class. The order of the questions seems to imply that you were supposed to use the tangential velocity to figure out the amount of energy, but I thought this way was easier for me. You can, and should, still try it that way to make sure the answers are the same. If they are not then one of us made a mistake... and I wouldn't be surprised if it was me.
705009
The total energy of the hoop as it enters the ramp is the sum of its forward kinetic energy and its rotational energy.  If it is rolling at w rad/sec, it is moving forward at a speed of 2*π*R*w.  The total kinetic energy is then .5*M*[2*π*R*w]^2 + .5*I*w^2   I is the moment of inertia of the hoop, M*R^2.\n\nThe total kinetic energy is converted to potential energy when the hoop stops.  Its potential energy is M*g*h, where h is the height of the hoop above the horizontal portion.  If the hoop travels a distance L up the ramp, it climbs an amount = L*sinø.  Now equating the potential energy at the top to kinetic energy at the bottom you get\n\n.5*M*[2*π*R*w]^2 + .5*M*R^2*w^2 = M*g*L*sinø\n\nYou have all the data to solve for L
538402
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.
1133901
Use the equation U = 0.5*k*x^2\n\nU = potential energy (joules)\nx = compression distance (meters)\nk = force constant (N/m)\n\nThis gives us the potential energy of the toy.\n\nU = 0.5 * 296 * 0.0122^2 = 0.0220 J\n\nThen calculate the Kinetic Energy:\n\nKE = 0.5 * m * V^2\n\nKE = Kinetic Energy (joules)\nm = Mass (kg)\nV = Velocity (m/s)\n\nKE = 0.5 * 0.153 * 0.299 = 0.0229 J\n\nTotal Energy = 0.0229 + 0.0220 = 0.0449 J\n\nFor part 2, assume total energy has been converted to potential energy and solve for compression distance:\nx = sqrt (2 * U / k) = sqrt (2 * 0.0449 / 296) = 0.0174 m\n\nAmplitude is twice the compression distance (for positive and negative)\nAmplitude = 2 * 0.0174 = 0.0348 meters\n\nFor part 3, assume total energy has been converted to Kinetic energy and solve for speed:\nV = sqrt (2 * KE / m) = sqrt (2 * 0.0449 / 0.153) = 0.766 m/s
458509
Conserve both momentum and total energy (i.e., potential energy plus kinetic energy).\n\nMy coordinate system will have the pendulum lifted to the left and the block on the right moving left, so that the initial momentum is - M2 Vo2.\n\nIt will also have the block at height 0, so the initial potential is M1 g h where h is the initial height of the pendulum, which looks like L(1 - sin(10 degrees)).\n\nThe initial kinetic energy is of course (1/2) M2 (Vo2)^2.\n\nSo ...\n\ninitial momentum: - M2 Vo2\ninitial energy: M1 g h + (1/2) M2 (Vo2)^2\n\nAfter the collision the potential energy has decreased to zero, and if w is the combined velocity of the two masses then ...\n\nfinal momentum: (M1 + M2)w\nfinal energy: (1/2)(M1 + M2)w^2 + [the energy "loss" from the slight heating and deformation of the two masses sticking].
794040
This is really a calc question.\n\nAn item will fall at 32ft per sec per sec from a rest.\nh - the free fall = t\n\n2. As it is stated the axle is frictionless, therefore the cyinder and axle will move and there is no force between the two. However there is a force between the rope and or bucket and the cylinder.\n\ndouble check the 32ft/s in your book, there may be more to the equation.\n\nCarl
755607
To solve this problem, you need to use conservation of energy.  At the beginning, the block is at rest at the top of this ramp.  At the bottom of the ramp, the block still has some kinetic energy, but some of it has dissipated due to friction.  Thus, the equation that governs this problem is:\n\nInitial Potential Energy = Dissipated Energy due to Friction - Final Kinetic Energy\n\nSince Potential Energy is defined only in the vertical direction, we must use the vertical components of the energy.\n\nWe can find the height of the ramp using trigonometry:\n\nsin(theta) = height/incline -----------> height = incline*sin(theta)\n\nheight = 3.03*sin(28.8) = 1.46 m\nPotential Energy = mgh = (2.51)(9.81)(1.46) = 35.95 J\nFrictional Energy (vertical) = u*m*g*sin(theta)*h\nKinetic Energy (vertical) = 0.5*m*[v*sin(theta)]^2 = (0.5)(2.51)(1.77sin(28.8))^2\n\nKinetic Energy = 0.91 J\n\nPlugging these values into the equation, you get:\n\n35.95 = u*m*g*sin(theta)*h + 0.91\n\nWe have all the values, now solve for u\n\n---------------\n\nHope this helps