- #1
IntegralBeing
- 5
- 0
Dear community, I am very ashamed to have not noticed this result but well into my physics major, and I would like to share it with you. I am not sure it is right, which is my reason for posting it.
The normal force is classically referred as
N = mg (1)
but if that was really the case; it is, if the Normal force completely canceled the gravitational force; then there would be no centripetal acceleration which were to keep a body on Earth's surface accelerating downwards to stay in contact with the planet.
Indeed, I think it is more like
N = k(mg) (2)
where k is the fraction of the gravitational force which does not contribute to the centripetal acceleration, which then is canceled by the Normal force.
Thus this centripetal acceleration is
ac = mv2/R (3)
with R being the radius of Earth, i.e. 6.37 × 106 m, and v the tangential velocity of a body on the surface being 469 m/s,
ac = 0.035 m/s2 (4)
then if we have
mg = N + Fc
= k(mg) + mac
then
k = (g - ac)/g = 0.997 (5)
First, this seems to be completely negligible, but it is a reason not to solve N = mg with more than 3 decimals of precision if one does not account for the centripetal acceleration of the studied body.
Second, I think it is a mistake not to introduce the normal force as N=mg-mac, mainly because of its theoretical significance. I, as an undergraduate, wasn't aware of this and I think it deserves minimal exposition at least in basic physics books. It is nice to have N = mg, but the normal force DOES NOT equal the gravitational force. It does not cancel it.
Am I wrong on this? I looked up the wikipedia article of normal force on Earth and it does not include the centripetal acceleration, which I think is wrong and misleading.
This is my first 'not very short' post; I am sorry for any errors! Please leave your comments.
The normal force is classically referred as
N = mg (1)
but if that was really the case; it is, if the Normal force completely canceled the gravitational force; then there would be no centripetal acceleration which were to keep a body on Earth's surface accelerating downwards to stay in contact with the planet.
Indeed, I think it is more like
N = k(mg) (2)
where k is the fraction of the gravitational force which does not contribute to the centripetal acceleration, which then is canceled by the Normal force.
Thus this centripetal acceleration is
ac = mv2/R (3)
with R being the radius of Earth, i.e. 6.37 × 106 m, and v the tangential velocity of a body on the surface being 469 m/s,
ac = 0.035 m/s2 (4)
then if we have
mg = N + Fc
= k(mg) + mac
then
k = (g - ac)/g = 0.997 (5)
First, this seems to be completely negligible, but it is a reason not to solve N = mg with more than 3 decimals of precision if one does not account for the centripetal acceleration of the studied body.
Second, I think it is a mistake not to introduce the normal force as N=mg-mac, mainly because of its theoretical significance. I, as an undergraduate, wasn't aware of this and I think it deserves minimal exposition at least in basic physics books. It is nice to have N = mg, but the normal force DOES NOT equal the gravitational force. It does not cancel it.
Am I wrong on this? I looked up the wikipedia article of normal force on Earth and it does not include the centripetal acceleration, which I think is wrong and misleading.
This is my first 'not very short' post; I am sorry for any errors! Please leave your comments.