A small disk with mass m and radius r sits at the very top of a larger half-circle disk with radius R.  The half-circle is flat on the ground (the diameter of the full-circle would be level with the ground).\n\nIf the smaller disk is nudged, at what point will the smaller disk lose touch with the larger disk?  There is no slipping, so the smaller disk will rotate as it travels down the slope of the larger disk.
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I'll give the framework of the answer, but without doing any of the actual work (math), since I'm on my own free time.\n   Define the key variable to be theta, the angle relative to vertical of the line from the center of the disk to the center of the half-circle.\n   Now, specify the formula for gravitational potential energy of the disk as a function of theta (OK, I'll do that: PE = mgh = mg (R+r) cos(theta) )\n   Now, specify the formula for kinetic energy of the disk as a function of d(theta)/dt.  This has two components, (1)energy associated with the velocity of the center of mass, v = d(theta)/dt * (R+r), [use KE=mv^2/2] and\n(2) energy associated with the spinning of the disk, omega = d(theta)/dt * (R/r + 1) [use KE= I(omega)^2 / 2] (the +1 in that equation is the "gotcha" in this problem)\n  Now, use conservation of energy to equate total kinetic energy to loss of gravitational potential energy.  This equation thus relates theta to d(theta)/dt.  By solving this differential equation with the proper initial conditions, you will get the function of theta versus time.\n   Determine the vertical acceleration by taking second time derivative of (r+R)cos(theta).  When this vertical acceleration equals g, the rolling disk loses touch with the large disk.
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r decreases becuase the ellipse is becoming thinner i think that would happen\n\nhope u find out if my answer is wrong GOOD LUCK!!!!!!!!
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dba_gregg has missed the point that both are same weight\n\nThe moment of Inertia of a hollow sphere  is 1.6 times greater than the solid sphere and so it rolls down slowly.\n\nThe speed of solid sphere will be 1.09 times the speed of hollow sphere.
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The moment of inertia of an object is given as:\n\nI = volume integral of (r² dm)\n\n(I have no good way of writing this, so check the wikipedia link below of more info). Wikipedia also has a list of moments of inertia, which I've included below. It provides the moment of inertia for a disk as:\n\nI = (1/2) * M * R²\n\nAs for calculating it step by step, note that if the mass is evenly distributed throughout the disk, then the density of the disk is equal to its mass divided by its volume:\n\nrho = M / (pi * R² * h)\n\nThe infintesimal dm can be expressed as:\n\ndm = rho * dV\n\ndm = M / (pi * R² * h) * r * d-theta * dr * h\n\ndm = M / (pi * R²) * r * d-theta * dr\n\nThus, the moment of inertia is given as:\n\nI = int(int(M / (pi * R²) * r^3, r=0..R),theta=0..2pi)\n\n(again, I don't have a good way to express this - its a double integral of M / (pi * R²) * r^3 * dr * d-theta, where r goes from 0 to R and theta goes from 0 to 2pi). This reduces to:\n\nI = int(M / (pi * R²) * r^3 * 2pi, r=0..R)\n\nI = M/(pi * R²) * (1/4)*R^4 * 2pi\n\nI = M/(R²) * (1/2)*R^4\n\nI = (1/2) * M * R²\n\nSo that's how you find the moment of inertia. Now, substituting your values into this expression provides:\n\nI = (1/2)(41 kg) (18.8 cm * 1 m/100 cm)²\n\nI = .7246 kg-m²\n\nThat's it! :)
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1. Usually you have to solve these problems in terms of energy, where initial energy = final energy + friction (lost) energy. The initial and final energies are each the sum of the respective initial and final potential + kinetic energies. (Sometimes the problem doesn't explicitly give you an initial potential energy; it may just tell you that an object slides x meters downhill.) If you know the initial and/or final momentum and either the mass or the velocity, you can find the missing variable (velocity or mass) and then transform the momentum into kinetic energy. \n2. The coefficient of friction 'mu' of course affects the friction force. Sometimes you are asked to find mu so you have to work backwards. You may know the lost energy and the length over which the object encountered friction. Then you solve for friction force. Then, knowing the normal force as well (force perpendicular to the friction surface), you can find mu.\n3. Inclines with friction are a big favorite of the problem writers. Just remember that the gravitational force normal to the incline is g*m*cos(theta) and the gravitational force along the incline is g*m*sin(theta). If a ramp is tilted just enough to allow the object to slide, g*m*sin(theta) = friction force = mu*g*m*cos(theta), so mu = sin(theta)/cos(theta) = tan(theta). Or if you know mu and want to find theta, it's arctan(mu).\n4. Another fave is centripetal force on a horizontal turntable with a friction surface. We know the centripetal force is w^2*r*m (w being angular rate in rad/sec). And we know the normal force is g*m. So, if we rev up the turntable until the object slides off, or we move the object out from the center until it slides off, we know that, at the instant it lets go, w^2*r=g*mu, so we can find w or r. Easy, right?\nI hope this helps; if you have followup questions I'll be watching this question until it closes.
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My layman's guess would be that the ring would start to roll earlier.  \n\nAssuming that the ring and the disc are the same thickness and made of the same material, the ring would be lighter, and that would mean it has less inertia and less force of friction to overcome.
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1)\nThis is a poorly-worded question.  You are probably asking for the acceleration, not the force, at that distance from the center.   The mass of the CD is not important in this context.  \nThe centripetal acceleration of points on the disc a distance r from the center is:\nac = v²/r\nWhere v is the linear velocity at that point.  You can find it like this:\nv = 2πr/T = 2π(0.2)/0.15 = 8.38 m/s\nSo\nac = (8.38)²/(0.2) = 350 m/s²\n*Note: you can't just multiply ac by the mass of the CD to get the 'force'.  It wouldn't mean anything.  But if you placed an object on the CD at this point, and multilpied ac by the OBJECT's mass, that would be the centripetal forced on the object.   \n\n2)\nF = m*v²/r ──► m = r*F / v²\nAgain, find the linear velocity, v, like this:\nv = 2πr/T = 2π(0.35)/(0.85) = 2.59 m/s\nSo\nm = (0.35)(3)/(2.59)² = 0.157 kg
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F - Ff = m*a\n\nThis is because the force you're applying is in one direction, while the frictional force is ALWAYS in the opposite direction. Frictional forces act to prevent the applied movement.\n\nBecause you're moving at constant velocity, you therefore have no acceleration, so:\n\nF - Ff = 0\n-or-\nF = Ff\n\nFf is the force due to friction which:\nFf = u * Fn\n\nWhere "u" (it's actually the greek letter mu, but this will do) is equal to the coefficient of friction, and Fn is equal to the perpendicular force acting on the box. On flat ground, the normal force is equal to the gravitational force, but on inclines you have to break it up into its vector components.\n\nF = u*Fg\nF = (0.20)*(25kg)*(9.8m/s^2)\nF = 49N\n\nYou see that the velocity at which you push it has no bearing on the answer because neither your applied force nor frictional forces are due to constant velocities.
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Your question contriducts itself.\n\nIt begins by stating that the child has a mass of 40 kg.  It ends with saying that we can neglect the childs mass and asks what the final angular velocity is.  Both these statements cant be true.\n\nThe idea in play here is the LAW of conservation of angular momentum.\n\nAngular momentum (L) = Rational Inertia (i&#92;I) * angula velocity (w, greek letter, lower case omega)\nL = w*I\n\nYou can probably treat the child as a point mass, by which we can neglect the volume/width of the child (NOT its mass) in order to calculate its rotational inertia.  You should be able to find a formula to calculate the rotational inertia of the disk (I dont have it memorized since it can be easily referenced).\n\nThe law of conservation of angular momentum states that angular momentum is conserved between the initial and final states of the system.\nInitially the child/rotating disk have some angular momentum.  After the child moves in towards the center, the system still have the same angular momentum, but now it is all in the disk the child doesnt have any.\n\nI am not going to do your homework for you, but surely you should be able to figure it out now.\n\n\nThere is an alternative solution.  In order for the child to move to the center of the rotating disk from the edge, the child had to do some work (work in the physical sense).  This work causes a change in rotational Kinetic energy of the disk and thus causes a change in angular velocity.  Find the amount of word the child did and solve for the change in angular velocity.\nrotational KE = 1/2 I*w^2\nBut the conservation of momentum method is the easier way to solve this problem I think.
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Liquid would hit the ground first due to the movement of the liquid inside the can. It produces more speed because the shift of gravity and would "beat" the other can to the base of the incline.\n\n---------------------------------------------\n\nIf anyone has other questions that they want sent specifically to me, please send them to yourqandaperson@yahoo.com -- Please keep them appropriate and ask only realistic questions or your question will be deleted and reported! Thank you for using this service!
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Angular momentum is conserved,\n\nAngular momentum is (inertia * angular velocity)\n\nInertia for a disk = .5*mass*radius*radius = 0.5 * 5.6 * .2 * .2\n\nThe initial angular momentum is then: 0.5 * 5.6 *.2 * .2 * 1.8 (note that is in rev/s)\n\nThe inertia for the clay is: 0.5 * 3.1 * .04 * .04\n\nThe final angular momentum of disk and clay is then: 0.5 * ((3.1 * .04 * .04) + (5.6 * .2 * .2)) * f   where f is the answer you are looking for.\n\nInitial angular momentum = final angular momentum:\n0.5 * 5.6 *.2 * .2 * 1.8 = 0.5 * ((3.1 * .04 * .04) + (5.6 * .2 * .2)) * f or\n\nf = (5.6 *.2 * .2 * 1.8)/((3.1 * .04 * .04) + (5.6 * .2 * .2)) in rev/sec
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Answer A) Small size & rocky surfaces
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Don't know what grade you're in, but looks like this is gonna involve some integration.\nLet the intersection point of the plane be z=h(=0.5 in this case).\n\nConsider a thin disc of thickness dz in the volume to be evaluated, at a distance z from the origin. Obviously, z>h. It would have a radius of \nr = sqrt(R^2 - z^2). R is the radius of the sphere, =1 in this case.\n\nHence, the volume of this disc would be\ndV = 2*pi*r^2*dz\n   = 2*pi*(R^2-z^2)*dz\n\nHence, for total volume, this is going to have to be integrated between z=h to z=R.\n\nI believe this works out to \nV = 2*pi*((2R^3/3)-R^2h+h^3/3) here. Put in the values mentioned above.\n\nWhew!
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5,25rad/s = 2πf\nf= 0,836Hz\nVt= (2πr*f)\nVt= 2π*0,0600m*0,836Hz= 0,315m/s\nAc= (V^2)/r\nAc= (0,315m/s)^2/0,0600m= 1,66m/s^2\n\nBoth points have the same frecuency, since they have the same angular velocity. So, for point 2 you already have a velocity of 5,25rad/s.\nThen, Ac= (V^2)/r →  (4π^2*r^2*f^2)/r→ (2πf)^2*r\nTherefore, Ac α r.\nIf r is half, then Ac is half (1,66m/s^2)/2= 0,83m/s^2\nthe same for the velocity.\n Vt= (2πr*f), because the radius is half and Vt α r\nVt= 2π*(r/2)*f = π*r*f= π*0,0600m*0,836Hz= 0,158 m/s\nOr just (0,315m/s)/2= 0,158m/s
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i got 100N to start the container moving and 80N to keep it moving at constant speed, but i'm not sure if i got it right. What is the answer on the back of your book? if it is the same as mine, i will help you.\n\nstatic friction is how "rough" the surface is between the base of the container and the ground.\n\nnot sure about this one but i think kinetic friction is how rough between a moving object and the ground.\n\nJust think of you are stand on the hill. The static friction keep you still but once you start to slide down, the friction is less.
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ok, so if we're assuming that it was a head on elastic collision, and all the energy was transferred then there are two possible situations: if the 25lb ball was sliding then the total energy transferred will be equal to the amount it had originally (since we assumed elastic collision) that is equal to 1/2mv^2...where mass would usually be measured using the metric system. \nif the ball was rolling then it has not only translational energy (1/2mv^2) but also rotational energy which is equal to 1/2Iw^2 where I = inertia of a ball...if it is a solid sphere the moment of inertia is equal to 2/5mr^2 (m = mass, r = radius) and w (omega) = rotational speed in radians per second...which is equal to velocity/radius. so the amount of energy transferred would be equal to 1/2Iw^2 + 1/2mv^2.
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Consider two objects having masses 1kg and 5 kg are in a horizontal plane.\n\nWe apply a force of 10 newton to the 1 kg mass and 50 newton to the 5kg mass. \n\nThat is, we apply less force for less mass and greater force for greater mass.\n\nNow we find the acceleration of the two bodies.\n\nThe1 kg mass will have an acceleration of 10/1 = 10 m/ s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nSimilarly the 5 kg mass will have an acceleration of 50/5 = 10 m/s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nBoth have the same acceleration and both have the same speed at any time after start.\n\nNow returning to gravitational pull, let us place the 1kg mass in one pan of the balance and the 5kg mass in another pan. \n\nThe pan having 5kg mass will come down and the pan having 1kg mass will go up.\n\nThis is because the earth is pulling the 5 kg mass with a force of 50 newton and 1kg mass with a force of 10 newton.\n\nIf  we drop the two objects from the same place, as we have seen already the 1 kg mass will fall down with an acceleration of 10 m/ s^2 and the 5kg mass will also fall down with 10 m/ s^2. But the forces acting on them are different.\n\nThe Earth’s pull is such that it gives less force for less mass and greater force for greater mass.\n\nTo be more specific it gives a force of 10 newton for every kilogram of mass and hence the acceleration is for any mass is the same as approximately 10m/s^2.