I would like an answer that a high school physics student would understand. Please don't include rotational physics. Thanks!
213342
At the top of the halfpipe, the ball is just sitting there with potential energy stored up (PE = mgh, where m = mass, g = acceleartion due to gravity, and h = height).  As the ball rolls downward, its PE lessens and its kinetic energy increases (KE = 1/2(mv^2), where v = velocity of the ball).  At the bottom of the halfpipe, all PE is gone and KE is at maximum.  As the ball goes up the other side, the KE decreases and the PE increases.  In a perfectly frictionless halfpipe, the ball would return to the same height as it was before it started rolling, and the KE would go to 0 and the PE would be at its maximum.\n\nIf you just dropped the ball (no halfpipe - just a freefall), the KE just before impact would be the same as the PE before dropping.  The two are inversely proportional - as one goes down, the other goes up.\n\nNow if you knew the exact shape of the halfpipe, you could start plugging in numbers to find the PE and KE of the ball at various points.
456243
Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
252254
Well it can be explained with a simple example take two coins ,fix one of the with your hand and rotate the other on the surface of the first coin .You will see that coin will have complete revolution in 180 degrees but in ordinary situations 360 degrees = one complete rotation .This new phenomenon is because of the property of the surface of the first coin. For explaining the spin of electron which takes 720 for complete revolution can be explained by a ballarina spinning on her toe while rotating a wine glass in her hand simeltaniously.
384767
(a) Assuming the base is a half-circle and that there are no other forces acting on the mass (friction, drag, etc), then the kinetic energy at the base is equal to the potential energy at the top. Note that the change in potential energy is not affected by the semi-circular path the mass follows, but rather the net change in height. This net change is simply the radius of the half-circle container:\n\nKE = mgR = (.71 g * 1 kg/1000 g)(9.8 m/s²)(25 cm * 1 m/100 cm)\n\nKE = .00174 J\n\n(b) The kinetic energy associated with rotation is given as:\n\nKER = (1/2) I w², where w is the rotational velocity. Note that since the sphere rolls without slipping, the translational velocity of its cener of mass is simply:\n\nv = w * r\n\nSubstituting this into the expression for the rotational kinetic energy:\n\nKER = (1/2) (I / r²) v²\n\nNote that the expression for the translational kinetic energy is:\n\nKET = (1/2) m v²\n\nThus, the total kinetic energy of the ball at the base of the hemisphere is;\n\nKE = KER + KET = (v²/2)( I/r² + m)\n\nThe fraction of the rotational kinetic energy over the total is:\n\nf = KER / KE = (v²/2)( I/r²) / (v²/2)( I/r² + m)\n\nf = ( I/r²)/( I/r² + m)\n\nFinally, the moment of inertia, I, of a sphere is:\n\nI = (2/5)mr²\n\nSubstituting this into the expression for f:\n\nf = (2/5)m/( (2/5)m + m)\n\nf = (2/5)/( (2/5) + 1)\n\nf = 2/7\n\n(c) At the base of the hemisphere, there are two forces acting on the ball: the force of gravity and the centripetal force. At the base, these forces are coincident (act in the same direction). Thus the normal force, which opposes these two, is simply:\n\nN = Fgrav + Fcent\n\nN = mg + mv²/R\n\nTo find v, recall that KET / KE is 2/7. Using this, we can manipulate the terms to get v:\n\n(1/2) (I / r) v² = (2/7)(mgR)\n\n(1/2) ((2/5)m) v² = (2/7)(mgR)\n\n(1/5) v² = (2/7)(gR)\n\nv = sqrt((10/7)(gR))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = 1.871 m/s\n\nFinally, the normal force is computed as:\n\nN = mg + mv²/R\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = .0169 Newtons\n\n\nGood Luck! :)
531381
Hi Yo-Yo!\nEnergy conservation law: m*g*h=E1+E2, m=0.12kg, h=1.2m, g=9.8m/s^2, E1 is translation kinetic energy, E2 is rotational kinetic energy.\nE1=m*v^2/2, v is translation speed at full length h, E2=I*w^2/2, w=angular speed at full length h.\nNow v=w*r, r=3.2*10^(-3) m, thus m*g*h=m*(w*r)^2/2+I*w^2/2, hence w=sqrt(2*m*g*h/(m*r^2+I)). Plug numbers in and find w & v (be a good boy!).\nMeanwhile v=a*t & h=a*t^2/2, ‘a’ is linear acceleration.\nThus t=v/a & h=a*(v/a)^2/2, so a=v^2/(2*h). Plug numbers in and find a & t. Angular acceleration being w’=a/r. My regards!
604454
I don't know but, I suggest you put this in the "Physics" category if you want the best possible answer.
573459
there is friction in space as said before but it is also because there is matter in space, just not very much, and eventually it would cause the object to stpo spinning
587823
Has anyone notices that the answer given by "d_1167" has absolutely nothing to do with the question which was asked?\nIn fact, his answer was not even his own, he merely copy-pasted it from a web site and then did not even have the decency to site it,\nhttp://education.vsnl.com/karnik/velocity_of_light.pdf\nor the HTML version,\nhttp://66.218.69.11/search/cache?p=education.vsnl.com%2Fkarnik%2Fvelocity_of_light.pdf&ei=UTF-8&fr=FP-tab-web-t&x=wrt&u=education.vsnl.com/karnik/velocity_of_light.pdf&d=FNtzZDmtMwG4&icp=1&.intl=us\n\n\nAnyway, to get back to the question at hand,\nI would have to lean somewhat towards the direction of the answer given by "NotEasilyFooled".\nThe speed at which the effects of cutting the rope on planet M would not reach the 2nd planet (N) until at least 1 year after it happened since no information can travel fast than light.\n\nSince the rope is massless (but we assume it to act as a normal rope anyway), there is no reason to think that it would necesarily fall the the surface of of planet N except for the fac that there was a force of tension in the rope when it was cut, pulling it towards the surface of planet N.  But we already assumed it was massless (or at least "weigthless")....applying a force to a massless object puts us in a rather awkward position (F = ma).
222262
Sounds like a Math/Science type question to me. Why are you posting it here?\n\nBesides, I have the the slightest idea.
291422
Go to your local physics professor. If you are correct, (s)he will no doubt help you to get it published. If you are incorrect, (s)he should be willing to show you where your math is off.
9915
ok.  call the downward translational velocity of the whole thing v.  this means that the tangential velocity at the edge of the poles (at the radius r) is equal to v.  the reason for this is that the strings are wound AROUND THE POLES, so this winding and the condition that the string has to unwrap for the thing to move, sets the angular velocity w = v/r.\n\nnow, if you look at the main wheel part, the angular velocity has to be equal to the angular velocity of the poles, since they are one solid piece.  however, as you note the radius of the wheel R > r.  so, the fact that v / r = w = V / R implies that V > v.  in other words, the motion of the outer edges of the wheel is faster than the motion of the outer edges of the rods.\n\ni think you are getting confused by saying that V is the translational velocity of the wheel.  it's not.  v is the translational velocity of the wheel, and of the rods.  the fact that the string is wound *around the rods* and not around the wheel sets the angular velocity of the whole apparatus equal to v/r.  V, the tangential velocity of the outer edges of the wheel, is much faster than the translational velocity.\n\nthat clear?
395434
The problem is about conservation of angular momentum.  The tricky part is (b). To solve this use this definition of angular momentum and draw a picture to find out the value of r sin θ\n\nThe angular momentum of a particle of mass m with respect to a chosen origin is given by vector product L = r x p= mvr sin θ
147920
Wow, that is a very good and tricky question. If you really want to know please send me an e-mail and I will tell you the most recent answer proposed by the science community.
259254
The angular momentum is conserved, and linear is not.\n\nGravity is a central force (meaning one that only depends on the radial distance between objects) and central forces conserve angular momentum.  \n\nA way to think of it is this:  angular momentum is something like mass x radius x linear velocity.  Watch an orbit simulation and you'll notice that as earth gets closer to the sun (radius decreases) it's speed increases.  Also as the earth gets farther from the sun, it slows down.  The two trade off (i.e. increase radius, decrease linear velocity, and vice versa) in a way that their product stays constant.  \n\nThe linear momentum is not constant for the same reason.  Linear momentum is mass x velocity, and from the above it's easy to see that velocity is not constant (there is one wrinkle:  if the orbit is circular then the magnitude of the velocity is constant, but its direction is not; so the velocity, a vector quantity that has magnitude and direction, is changing).\n\n\nI hope that helps.
424569
Oscar. You have written too much. This 125m tower stuff does not matter compared to r=6.38*10^6. Why is this talk about 1.00 km and angle theta? See what “zipper” above says. Besides I answer your last question << Why is the trajectory of the cannonball not parabolic?>>. Why should it be if both you and Newton demand it to be circular? Assume that V1 of cannonball is the 1st cosmic speed corresponding to a circular track. Let V2 be > than V1, then you receive an oval track. Take V3 > than V2, then the track may become parabolic, so that cannonball would never return back to the cannon to linger out among planets and stars. If you are intrigued calculate a set of velocities and corresponding orbits of the cannonball. If cannon is not available consider a football! I like your inventive wits – to be honest!
19029
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!
57573
The Earth was formed spinning (like a top) and it just hasn't stopped spinning since then.  There is very little friction that would slow it down (space is pretty empty).\n\nThere are different theories as to why it was created spinning.  But most likely it formed from gas and dust that was swirling, and that angular momentum is still with us today.