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g s
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There's nothing wrong with JEB's answer, but I think this may help:

You write

Now I know that centrifugal and centripetal force are opposite to each other 
...
But then why... the rotating object faces a centrifugal force outwards?

Centrifugal and centripetal forces always point in opposite directions, along the radius away from or towards the center of rotation. However, they need not be equal in magnitude. The expression you found for centripetal force is only identical to the expression for centrifugal force (except for the direction) for the special case of an object for which there is no acceleration in the radial direction in the frame which measures the centrifugal force. In other cases, the $r$ means something different in the two expressions. See the note at the end if interested.

To explain in more detail:

Let us take the frame of an observer who is rotating with angular velocity $\omega$ about the point $r=0$. For instance, we can attach the observer to a big rotating wheel.

If we attach a relaxed spring to a weight on one end and to the spinning wheel on the other end, with initial velocity 0 in the radial direction, as measured from a frame corotating with the wheel, we will start out by measuring a centrifugal force on the weight and zero centripetal force. The weight accelerates in the radial direction. This gives it a velocity in the radial direction, which makes it increase in radius. As it increases in radius, the inward pointing (centripetal) force of the stretched spring increases.

Eventually, the system will reach an equilibrium in which there is no longer any acceleration in the radial direction. This will be the amount of stretch at which the centripetal force equals the centrifugal force in the co-rotating frame.

If the body is accelerating in the co-rotating frame, the $r$ in the canonical expression for centripetal force does not indicate the distance to the center of rotation of the co-rotating frame, but the distance to a point at which the velocity of the body would be the tangent velocity of a circular path. This is the radius of curvature. If we have a free body travelling in a straight line, the radius of curvature is infinite, because only in the limit as radius goes to infinity does the circumference of a circle approach a straight line. Note how $F=mv^2/r = 0$ in the limit as $r \to \infty$. If the object is accelerated to follow a curved path, $r$ becomes finite. The tighter the curve, the smaller the radius of curvature. In the case for which the object is accelerated to follow a curved path corotating with the observer who measures the centrifugal force, the radius of curvature equals the radius from the observer's center of rotation to the object under consideration. In all other cases, the displacement from the center of rotation to the rotating object is not the radius of curvature.

There's nothing wrong with JEB's answer, but I think this may help:

You write

Now I know that centrifugal and centripetal force are opposite to each other 
...
But then why... the rotating object faces a centrifugal force outwards?

Centrifugal and centripetal forces always point in opposite directions, along the radius away from or towards the center of rotation. However, they need not be equal in magnitude. The expression you found for centripetal force is only identical to the expression for centrifugal force (except for the direction) for the special case of an object for which there is no acceleration in the radial direction in the frame which measures the centrifugal force. In other cases, the $r$ means something different in the two expressions. See the note at the end if interested.

To explain in more detail:

Let us take the frame of an observer who is rotating with angular velocity $\omega$ about the point $r=0$. For instance, we can attach the observer to a big rotating wheel.

If we attach a relaxed spring to a weight on one end and to the spinning wheel on the other end, with initial velocity 0 in the radial direction, as measured from a frame corotating with the wheel, we will start out by measuring a centrifugal force on the weight and zero centripetal force. The weight accelerates in the radial direction. This gives it a velocity in the radial direction, which makes it increase in radius. As it increases in radius, the inward pointing (centripetal) force of the stretched spring increases.

Eventually, the system will reach an equilibrium in which there is no longer any acceleration in the radial direction. This will be the amount of stretch at which the centripetal force equals the centrifugal force in the co-rotating frame.

If the body is accelerating in the co-rotating frame, the $r$ in the canonical expression for centripetal force does not indicate the distance to the center of rotation of the co-rotating frame.

There's nothing wrong with JEB's answer, but I think this may help:

You write

Now I know that centrifugal and centripetal force are opposite to each other 
...
But then why... the rotating object faces a centrifugal force outwards?

Centrifugal and centripetal forces always point in opposite directions, along the radius away from or towards the center of rotation. However, they need not be equal in magnitude. The expression you found for centripetal force is only identical to the expression for centrifugal force (except for the direction) for the special case of an object for which there is no acceleration in the radial direction in the frame which measures the centrifugal force. In other cases, the $r$ means something different in the two expressions. See the note at the end if interested.

To explain in more detail:

Let us take the frame of an observer who is rotating with angular velocity $\omega$. For instance, we can attach the observer to a big rotating wheel.

If we attach a relaxed spring to a weight on one end and to the spinning wheel on the other end, with initial velocity 0 in the radial direction, as measured from a frame corotating with the wheel, we will start out by measuring a centrifugal force on the weight and zero centripetal force. The weight accelerates in the radial direction. This gives it a velocity in the radial direction, which makes it increase in radius. As it increases in radius, the inward pointing (centripetal) force of the stretched spring increases.

Eventually, the system will reach an equilibrium in which there is no longer any acceleration in the radial direction. This will be the amount of stretch at which the centripetal force equals the centrifugal force in the co-rotating frame.

If the body is accelerating in the co-rotating frame, the $r$ in the canonical expression for centripetal force does not indicate the distance to the center of rotation of the co-rotating frame, but the distance to a point at which the velocity of the body would be the tangent velocity of a circular path. This is the radius of curvature. If we have a free body travelling in a straight line, the radius of curvature is infinite, because only in the limit as radius goes to infinity does the circumference of a circle approach a straight line. Note how $F=mv^2/r = 0$ in the limit as $r \to \infty$. If the object is accelerated to follow a curved path, $r$ becomes finite. The tighter the curve, the smaller the radius of curvature. In the case for which the object is accelerated to follow a curved path corotating with the observer who measures the centrifugal force, the radius of curvature equals the radius from the observer's center of rotation to the object under consideration. In all other cases, the displacement from the center of rotation to the rotating object is not the radius of curvature.

deleted 344 characters in body
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g s
g s
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There's nothing wrong with JEB's answer, but I think this may help:

You write

Now I know that centrifugal and centripetal force are opposite to each other 
...
But then why... the rotating object faces a centrifugal force outwards?

Centrifugal and centripetal forces always point in opposite directions, along the radius away from or towards the center of rotation. However, they need not be equal in magnitude. The expression you found for centripetal force is only identical to the expression for centrifugal force (except for the direction) for the special case of an object for which there is no acceleration in the radial direction in the frame which measures the centrifugal force. In other cases, the $r$ means something different in the two expressions. See the note at the end if interested.

To explain in more detail:

Let us take the frame of an observer who will measure a centrifugal force matching the canonical formula. This observer has the sameis rotating with angular velocity $\omega$ about the same center of rotation as the object we are considering. That is, the observer and the object arepoint co-rotating$r=0$. For instance, on a spinning wheel, all points on the wheel are co-rotating, as are all points on an imaginary larger wheel that we could make by expandingcan attach the wheel, no matter howobserver to a big we made it. But points on another wheel on the same car are not co-rotating, even though they have the same angular velocity, because they are rotating about a different axle.

In this frame, if we have a bigger centrifugal force than a centripetal force, we will measure it to accelerate in the radial direction, in conformity with $F=ma$, and when centripetal force is greater, we will measure it to accelerate opposite the radial directionwheel.

For example, ifIf we attach a relaxed spring to a weight on one end and to the spinning wheel on the other end, with initial velocity 0 in the radial direction, as measured from a frame corotating with the wheel, we will start out by measuring a centrifugal force on the weight and nozero centripetal force. The weight accelerates in the radial direction. This gives it a velocity in the radial direction, which makes it increase in radius. As it increases in radius, the inward pointing (centripetal) force of the stretched spring increases.

Eventually, the system will reach an equilibrium in which there is no longer any acceleration in the radial direction. This will be the amount of stretch at which the centripetal force equals the centrifugal force in the co-rotating frame.

If the body is accelerating in the co-rotating frame, the $r$ in the canonical expression for centripetal force does not indicate the distance to the center of rotation of the co-rotating frame.

There's nothing wrong with JEB's answer, but I think this may help:

You write

Now I know that centrifugal and centripetal force are opposite to each other 
...
But then why... the rotating object faces a centrifugal force outwards?

Centrifugal and centripetal forces always point in opposite directions, along the radius away from or towards the center of rotation. However, they need not be equal in magnitude. The expression you found for centripetal force is for the special case of an object for which there is no acceleration in the radial direction in the frame which measures the centrifugal force.

To explain in more detail:

Let us take the frame of an observer who will measure a centrifugal force matching the canonical formula. This observer has the same angular velocity about the same center of rotation as the object we are considering. That is, the observer and the object are co-rotating. For instance, on a spinning wheel, all points on the wheel are co-rotating, as are all points on an imaginary larger wheel that we could make by expanding the wheel, no matter how big we made it. But points on another wheel on the same car are not co-rotating, even though they have the same angular velocity, because they are rotating about a different axle.

In this frame, if we have a bigger centrifugal force than a centripetal force, we will measure it to accelerate in the radial direction, in conformity with $F=ma$, and when centripetal force is greater, we will measure it to accelerate opposite the radial direction.

For example, if we attach a relaxed spring to a weight on one end and to the spinning wheel on the other end, with initial velocity 0 in the radial direction, as measured from a frame corotating with the wheel, we will start out by measuring a centrifugal force on the weight and no centripetal force. The weight accelerates in the radial direction. This gives it a velocity in the radial direction, which makes it increase in radius. As it increases in radius, the inward pointing (centripetal) force of the stretched spring increases.

Eventually, the system will reach an equilibrium in which there is no longer any acceleration in the radial direction. This will be the amount of stretch at which the centripetal force equals the centrifugal force in the co-rotating frame.

There's nothing wrong with JEB's answer, but I think this may help:

You write

Now I know that centrifugal and centripetal force are opposite to each other 
...
But then why... the rotating object faces a centrifugal force outwards?

Centrifugal and centripetal forces always point in opposite directions, along the radius away from or towards the center of rotation. However, they need not be equal in magnitude. The expression you found for centripetal force is only identical to the expression for centrifugal force (except for the direction) for the special case of an object for which there is no acceleration in the radial direction in the frame which measures the centrifugal force. In other cases, the $r$ means something different in the two expressions. See the note at the end if interested.

To explain in more detail:

Let us take the frame of an observer who is rotating with angular velocity $\omega$ about the point $r=0$. For instance, we can attach the observer to a big rotating wheel.

If we attach a relaxed spring to a weight on one end and to the spinning wheel on the other end, with initial velocity 0 in the radial direction, as measured from a frame corotating with the wheel, we will start out by measuring a centrifugal force on the weight and zero centripetal force. The weight accelerates in the radial direction. This gives it a velocity in the radial direction, which makes it increase in radius. As it increases in radius, the inward pointing (centripetal) force of the stretched spring increases.

Eventually, the system will reach an equilibrium in which there is no longer any acceleration in the radial direction. This will be the amount of stretch at which the centripetal force equals the centrifugal force in the co-rotating frame.

If the body is accelerating in the co-rotating frame, the $r$ in the canonical expression for centripetal force does not indicate the distance to the center of rotation of the co-rotating frame.

Post Deleted by g s
Source Link
g s
g s
  • 14.1k
  • 2
  • 21
  • 43

There's nothing wrong with JEB's answer, but I think this may help:

You write

Now I know that centrifugal and centripetal force are opposite to each other 
...
But then why... the rotating object faces a centrifugal force outwards?

Centrifugal and centripetal forces always point in opposite directions, along the radius away from or towards the center of rotation. However, they need not be equal in magnitude. The expression you found for centripetal force is for the special case of an object for which there is no acceleration in the radial direction in the frame which measures the centrifugal force.

To explain in more detail:

Let us take the frame of an observer who will measure a centrifugal force matching the canonical formula. This observer has the same angular velocity about the same center of rotation as the object we are considering. That is, the observer and the object are co-rotating. For instance, on a spinning wheel, all points on the wheel are co-rotating, as are all points on an imaginary larger wheel that we could make by expanding the wheel, no matter how big we made it. But points on another wheel on the same car are not co-rotating, even though they have the same angular velocity, because they are rotating about a different axle.

In this frame, if we have a bigger centrifugal force than a centripetal force, we will measure it to accelerate in the radial direction, in conformity with $F=ma$, and when centripetal force is greater, we will measure it to accelerate opposite the radial direction.

For example, if we attach a relaxed spring to a weight on one end and to the spinning wheel on the other end, with initial velocity 0 in the radial direction, as measured from a frame corotating with the wheel, we will start out by measuring a centrifugal force on the weight and no centripetal force. The weight accelerates in the radial direction. This gives it a velocity in the radial direction, which makes it increase in radius. As it increases in radius, the inward pointing (centripetal) force of the stretched spring increases.

Eventually, the system will reach an equilibrium in which there is no longer any acceleration in the radial direction. This will be the amount of stretch at which the centripetal force equals the centrifugal force in the co-rotating frame.