A question about apparent weight.

[agag1002]

How to find the apparent weight of objects in a certain latitude?
the formula of the apparent weight is R=mg-mrω^(2)cosθ?
thank you~

 

[RedX]

You just plug in the right values for the variables.

m is the mass of the object whose weight you want. g is the acceleration due to gravity which is 9.8 m/s². r is the radius of the earth. omega is angular velocity of the earth, which is 2*3.14 radians per 23 hours+56 minutes. theta is the latitude of the location you're measuring the weight.

 

[Quinzio]

How to find the apparent weight of objects in a certain latitude?
the formula of the apparent weight is R=mg-mrω^(2)cosθ?
thank you~

cosine is squared
about 0.9999mg at the equator
$$m(g-r\omega^2cos^2\theta)$$

 

[agag1002]

cosine is squared
about 0.9999mg at the equator
$$m(g-r\omega^2cos^2\theta)$$

How to prove this formula?

 

[epenguin]

Well, just to nudge this along, with apologies for the fumbling, I guess I see where that comes from though am a bit uneasy.

Take the case it is at the equator. On a stationary Earth force towards centre is mg. Now (at equator) on a rotating Earth an object appearing stationary to someone else on the surface is actually accelerating vertically towards Earth centre at rate rω² . Revise that bit if not secure, necessary for planetary motion etc.

Various ways to explain physically and this is the difficult bit, but roughly... Fall with the force and you feel no force. Freefall. You are a tiny bit in freefall as per above when you are apparently standing still on Earth equator so this acceleration has to be subtracted in the calc. from the total acceleration which you would experience but which, resisting it, you feel as gravity force. Those who know what I mean can maybe explain it better.

Now for the cos part, firstly away from the equator the distance from the axis of rotation is less than the Earth radius r: it is not r but r cos θ so that's one cos. But that's a (pseudo) force parallel to the plane of the equator. The 'force' towards Earth centre (i.e. vertical as we experience it) is just the resolved component of that so there's another cos θ factor, together giving you cos²θ.

H'mm, I see then the component perpendicular to that amounts to some leftover sideways pseudo-force there, let's see (thinks er sun rises in East :uhh: so Earth is going anticlockwise for us majority) that would be Northwards in N hemisphere and Southwards in S. I think it should be maximum at latitude 45° N or S

(Is this right? I don’t remember hearing about it. This North- or Southwards apparent small force should be measurable and have been measured, presumably involving satellites as quite difficult to imagine how to do by local measurement – e.g. you have to have some other criteria of 'vertical', of direction of Earth c of g than direction of apparent attraction, or the direction a weight hangs.)

 

[Redbelly98]

A force diagram should be used to solve this. Apparent weight is the normal force exerted by the ground on the mass -- this is the weight that a scale would read. The net force must equal the centripetal force, $m \omega ^2 r$, directed toward the Earth's rotation axis (not toward the Earth's center, except at the equator).

 

[RedX]

The formula should probably be:

R=mg-mrω^(2) (cosθ)^2

where you project the centrifugal force in the direction towards the center of the earth, and the other component of the centrifugal force (tangential to earth) is too small to bother with. Of course the normal force balances everything (due to Earth's oblate shape). If you insist on precision, then:

R^2=[mg-mrω^(2) (cosθ)^2]^2 + [mrω^(2) (cosθ)(sinθ)]^2

and just take the square root to get the apparent weight.

edit:

oops, quinzio already corrected the formula.