Imagine the earth spinning faster and faster and faster .....
At some stage your feet would lift off from the surface of the Earth - which force do you think would try to push you off and what force would try to keep you on the Earth?
Imagine the earth spinning faster and faster and faster .....
At some stage your feet would lift off from the surface of the Earth - which force do you think would try to push you off and what force would try to keep you on the Earth?
I am guessing gravity will try to keep u on the earth. But which force will push u off ??
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Apparently the gravity cancels out the centrifugal force when the earth rotates really fast. ?
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You still haven't defined "weightless". If you do this you can then say what conditions are required to produce it. The main problem with this topic is that there is generally not a clear understanding of what the word means.
You still haven't defined "weightless". If you do this you can then say what conditions are required to produce it. The main problem with this topic is that there is generally not a clear understanding of what the word means.
Hm.... Sorry abt that. Hm... what would happen to any object on this planet's equator if earth rotates faster and faster ? Would their gravitational force cancel out with the centrifugal force ? I am actually quite confused at the moment.. My mind is in a state if chaos at the moment ....
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You still haven't defined "weightless". If you do this you can then say what conditions are required to produce it. The main problem with this topic is that there is generally not a clear understanding of what the word means.
And yes under what condition It happens and why does it happen. ?
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Ok, so the earth is rotating with angular speed w, and the radius of the earth is r. Suppose we have our object sitting on the scales on the surface of the earth:
There are two forces acting on the object. The weight force and the force P which is the scales pushing up on the object. The value of P will be used to determine the reading on the scales.
In order for the object to keep on rotating with the earth, the net force on it must m*(w^2)*r which is the centripetal force.
Therefore, mg - P = mw^2r so P = mg - mw^2r so increasing the speed of rotation will decrease P and hence decrease the reading on the scales, so the object would appear to weigh less if you were to try to weigh it.
Ok, so the earth is rotating with angular speed w, and the radius of the earth is r. Suppose we have our object sitting on the scales on the surface of the earth:
There are two forces acting on the object. The weight force and the force P which is the scales pushing up on the object. The value of P will be used to determine the reading on the scales.
In order for the object to keep on rotating with the earth, the net force on it must m*(w^2)*r which is the centripetal force.
Therefore, mg - P = mw^2r so P = mg - mw^2r so increasing the speed of rotation will decrease P and hence decrease the reading on the scales, so the object would appear to weigh less if you were to try to weigh it.
Ok.. What is this force P ? Is it the reaction force? Sorry if this question is stupid. My mind is in a state of chaos at the moment ..
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Ok.. What is this force P ? Is it the reaction force? Sorry if this question is stupid. My mind is in a state of chaos at the moment ..
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Yes it is. It's also used by the scales to determine the reading to display (for instance if it was a spring-operated scales then increasing P would increase how much the spring is pushed/stretched and the reading can be determined from there)
Yes it is. It's also used by the scales to determine the reading to display (for instance if it was a spring-operated scales then increasing P would increase how much the spring is pushed/stretched and the reading can be determined from there)
I see. So increasing in speed rotation suggests increase in the frequency of the rotation and w increases because that P=mg- mrw^2, so that P decreases
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