why... wouldnt there be? the wheel WILL turn, that much is certain (turn reltive to the fixed point on the earth, that is,,,,
ohh, i see what you mean. its the wheels own angular momentum being conserved, not any from the earth being transferred to it, right? On 7/20/07, Harry Veeder <[EMAIL PROTECTED]> wrote:
On 20/7/2007 2:48 PM, leaking pen wrote: > on a more serious answer, if you are only taking the earths rotation > into account, then yes, i would think so. remember, becuase of the > tilt of teh earth, 90 degrees straight with teh ground would not be > the right orientation, me thinks. I'm ignoring the tilt of the Earth relative to the orbit it makes around the sun. > How do you measure the force involved in a transfer of angular > momentum like that? I'm not sure there is a _transfer_ of angular momentum...which is why I find this "thought experiment" intriguing. Harry > On 7/20/07, Harry Veeder <[EMAIL PROTECTED]> wrote: >> I made a drawing of the situation I was imagining: >> >> http://web.ncf.ca/eo200/world-ferriswheel.html >> >> A ferris wheel located on Earth's equator. Initially a brake prevents the >> wheel from turning. After the brake is released and assuming the axel of the >> wheel is frictionless, will the orientation of the wheel remain unchanged as >> the Earth revolves? >> >> Harry >> >> >> On 20/7/2007 5:42 AM, Michel Jullian wrote: >> >>> Lets' assume the wheel axis is parallel to the Earth's for simplicity. >>> Moving >>> the hub around in the plane of the wheel shouldn't change its rate of >>> rotation >>> wrt an inertial frame if there is no friction. >>> >>> If it's initially non-rotating wrt the distant stars (i.e. rotating once per >>> day wrt the Earth) it will remain so, you can verify that with a bicycle, >>> lift >>> the front you'll see that the front wheel keeps the same orientation. >>> >>> If it's initially non-rotating wrt the Earth, i.e. rotating once per day wrt >>> the distant stars, it will keep its angular speed i.e. it will remain at >>> rest >>> wrt the Earth. >>> >>> Michel >>> >>> ----- Original Message ----- >>> From: "Harry Veeder" <[EMAIL PROTECTED]> >>> To: <[email protected]> >>> Sent: Friday, July 20, 2007 5:23 AM >>> Subject: Re: [Vo]:centripetal force question >>> >>> >>>> >>>> Imagine a ferris wheel (absent the carriages for simplicity) which is >>>> initially at rest. >>>> >>>> As the world turns, the ferris wheel will complete one revolution in a day >>>> (assuming no friction) with respect to an observer standing >>>> beside it. >>>> >>>> Yes? No? ...or? >>>> >>>> Harry >>>> >>>> >>>> >>>> >>>> On 15/7/2007 5:17 PM, Michel Jullian wrote: >>>> >>>>> Hi Thomas, >>>>> >>>>> The (fictitious, or apparent) force you're talking about is a function of >>>>> _rotations_ (not revolutions) per second, and also of your mass and of >>>>> your >>>>> distance from the axis >>>>> (force=mass*(2*pi*rotations_per_second)^2*radius_of_the_earth assuming you >>>>> stand on the equator), nothing to do with distance traveled by the planet, >>>>> and >>>>> it is not centripetal (going towards the center) but centrifugal (think of >>>>> fugitive = going away from the center), if it was centripetal it would not >>>>> subtract from but add to actual weight, which is the actual centripetal >>>>> force. >>>>> >>>>> Michel >>>>> >>>>> ----- Original Message ----- >>>>> From: "thomas malloy" <[EMAIL PROTECTED]> >>>>> To: <[email protected]> >>>>> Sent: Sunday, July 15, 2007 10:02 PM >>>>> Subject: [Vo]:centripital force question >>>>> >>>>> >>>>>> I'm subject to weight loss produced by the centripital force produced by >>>>>> the earth's rotation, I'm wondering if centripetal force is a function >>>>>> of revolutions per time unit, or total distance traveled as the planet >>>>>> travels? >>>>>> >>>>>> >>>>>> --- http://USFamily.Net/dialup.html - $8.25/mo! -- >>>>>> http://www.usfamily.net/dsl.html - $19.99/mo! --- >>>>>> >>>>> >>>> >>> >> >> >
-- That which yields isn't always weak.
