Physics Help Forum centrifugal force and net weight

 Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum

 Jan 18th 2015, 06:41 AM #1 Junior Member   Join Date: Jan 2015 Posts: 15 centrifugal force and net weight When an object, orbiting in circular path, experiences the sensation of being thrown outward away from the centre of circle. We often think that an outward force or centrifugal force is responsible for this trend. But it is wrong idea. Inertia is responsible for this. Since earth rotates around its axis, an object standing on the surface of earth rotates with earth. We say that it affect the downward force that object feels while standing on the surface of earth, and to calculate the net value of attraction force between earth and object we subtract the value of centrifugal force from the weight of that object. This is because centrifugal force pulls this object outward, and the value of centrifugal force equals to that of centripetal force. But we know that centrifugal force is not responsible for the sensation of being thrown from the centre of circle. So why, in the case of effect of earth's rotation on value of gravitational force, do we think that centrifugal force wants to pull the object to outward?
 Jan 18th 2015, 08:10 AM #2 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,271 As you point out - there is no such thing as centrifugal force. There is centripetal acceleration, equal to v^2/r, and the force required to cause that acceleration is the centripetal force F = mv^2/r. The net weight of an object on the Earth's surface is the sum of the gravitational force it experiences minus the portion of gravitational force required to maintain that centripetal acceleration. Hence: W = m(gh-v^2/r) where v is the velocity of the point on the surface of the Earth due to rotation and r is the radius of the Earth. The magnitude of that second term is much less than the first term, so in general we ignore it.

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