Physics Conservation of Angular Momentum?
952670
Conservation of angular momentum must be considered as PhysicsDude mentioned. The thing to keep in mind here is the speed of the train given is "with respect to the track". The train is moving in one direction and the wheel/track is the other. If the wheel were not moving the speed of the train would be the 0.15 m/s, however since the wheel is moving counter to the train, and observer standing by the track (that is moving on the wheel along with the track) would see the speed of the train as faster than the actual train speed by an amount equal to the speed of the track. \n\nVr = Vtt + Vtr\n\nwhere Vr (=0.15 m/s) is the relative speed, Vtt is the speed of the toy train, and Vtr is the speed of the track.\n\nNow, considering conservation of angular momentum\n\nLtt = Itt*wtt = Itr*wtr = Ltr\n\nwhere Itt is the moment of inertia of the train, wtt is the angular speed of the train, Itr is the moment of inertia of the track, and wtr is the speed of the track. The two equate since the intial angular momentum is zero so that the two must cancel each other out after the train has started moving. The moments of inertia are:\n\nItt = m*r^2\nItr = 1.1*m*r^2\n\nwhere m is the mass of the train (1.1*m is the mass of the wheel/track), and r is the radius of the wheel. Recall that the angular speed is the linear speed divided by the radius. This means that the above conservation equation is\n\nm*r^2*wtt = m*r^2*(Vr/r - wtr) = 1.1*m*r^2*wtr\n\nSolving for the angular speed of the wheel/track (wtr)\n\nwtr = Vr/(2.1*r)\n\nPlugging in the values given\n\nwtr = (0.15 m/s)/(2.1*0.43 m) = (0.15 m/s)/(0.903 m)\n\nwtr = 0.166 rad/s
824196
conservation of linear moment:\nm1*v1 = m2*v2\nm1*v1/m2 = v2\n(100.0 kg)(3.0 m/s)/(175.0 kg) = v2 = 1.71 m/s
914874
yes if it is spinning on its own axis as it moves.  This is called Spin Angular Momentum
611220
This is about conservation of momentum. Any object on a rotating (and for our experiment atmosphere-less) planet has the same rotational momentum and angular velocity  as the surface. If it jups up, it will increase the radius, and therefore reduce the angular velocity, so it will land further behind.
523453
Why do you think rotation isn't a natural motion? Angular momentum is conserved, so when a gravitational collapse happens, the natural tendency *is* to rotate. The only exception would be a situation when everything was initially at rest (which wasn't the case). Also, why do you think all the stars rotate *in sync*? Each one is going in its own orbit with its own velocity. These velocities are not all the same by any means.
998322
Lets make an enormous assumption to fill in the gaps.\nYou have a thing on a string of length L and spin it round at a certain rate of rotation. Keeping the rate of roataion what would the tangential velocity of the thing be if the string was shortened to L/2?\nIn this case the tangential velocity (= the radius x the angular velocity) will also halve.
710256
You got the CG just right.  Your method might work with enough concentration., but I would look at making the angular momentum come to zero:\nL = Iω = (2.7*.325²  + 1.3*.675²) * 26rpm*2π/60 = 2.389 kg m²/s\n\nTo stop the movement the angular impulse must equal the - of the angular momentum:\n\nT * t = L→ T = L/t = 2.389/5.8 = .4118 Nm (ccw)
909738
It will rotate about the center of mass O at d/2\ncalculate the moment or inertia about  O, include both mass and both rods. see link for formulas if you don't know how\nhttp://en.wikipedia.org/wiki/Moment_of_inertia\nThe formula for KE is also there
988688
linear momentum of a moving body is the "thing" that keeps it moving, given by mass x velocity.\n\nangular momentum is the same "thing" that keeps an object spinning about its axis.
256338
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.
818010
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
816105
Generally, some of their collective kinetic energy is converted into heat, so the KE after the collision is less than the initial KE.  Momentum is conserved.
742331
YES,\n\nand I dont even need to do the calculations. \n\nPHYSICS laws state that momentum will stay constant, so is the energy.
943735
a) T = dL / dt = 16\nL = 16 t \n\nI don't think you can do any better than that since they don't ask for the angular momentum at a particular moment in time, and it certainly is not constant.\n\nb) L = Iw\nw = L / I = 16 t / I = 16(33 * 10^-3) / (1.2 * 10^-3) = 440 rad/s
809093
Angular momentum=moment of inertia* angular velocity\nconservation of angular momentum applied to before and after the move.
1165448
One earlier answer is close but does not go all the way.  \n\nThe material it formed out of was slowly rotating already, and angular momentum is conserved;  the more compact that material became the faster it had to rotate to conserve angular momentum.\n\nWhy did the material have some rotation in the first place?  Because it is statistically VERY improbable to have a large extended ANYTHING with net zero rotation.  Contract such an anything enough and its rotation becomes appreciable.\n\nIt is worth noting that it is the combined angular momentum of the Earth-Moon system that is relevant here. The moon once orbited much closer and the Earth rotated much faster.  The transfer between the two (terrestrial rotation and the moon's orbital motion) continues, but the total of the combined has changed little since the hypothesised collision that formed the moon.
1071120
Stars and planets form as a result of the gravitational collapse of accreting material. Any net translational motion of that initial material is accelerated as its radius of rotation decreases. This is due to the same conservation of angular momentum principle that makes skaters and divers spin more rapidly when they bring their arms closer to their bodies. In its extreme, this effect can make collapsed neutron starts spin many times each second, creating the regular beat of pulsars as observed from Earth. \n\nOnce an initial spin was established for the Earth, the same conservation principle says that its angular momentum will continue unless it can be transferred to another object. With no significant frictional forces to allow that to happen, the Earth can continue to spin indefinitely without any further prodding. (More on that later.) \n\nIt is incorrect to say that the Moon does not spin. It rotates on its axis in synch with its revolution around the Earth, keeping the same face always pointed toward us. If it did not rotate at all, we would see all of its surface over the course of one month's orbit. This condition was created by tidal forces between the Earth and Moon, gradually slowing its initial rotation until the current more stable situation evolved. This same 'tidal lock' condition has been observed for satellite closely orbiting other planets. Because of its elliptical orbit, a slight 'wobble' allows us to actually see slightly more than half of the Moon's surface over the course of a month. \n\nTidal forces, by the way, continue to affect the Earth/Moon system. They exert enough frictional force on the Earth so that some of its angular momentum IS being transferred slowly to the Moon. The net effect of that is to gradually slow down Earth's rotation and lengthen the day. Where is that energy and momentum going? The moon is gradually receding from the Earth. Both of these effects are small. The Moon is receding at a rate of less than 2 inches per year and the day is getting longer by one second every 67,000 years or so.