Why does radial acceleration act toward the center?

In summary, centrifugal acceleration is the acceleration of an object that is moving in a circular path and is directed away from the center of the circle.
  • #1
Athi Sankar
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Acceleration of a rotating link has two components,Tangential (change in the direction) Radial (change in the magnitude). Why the direction of Radial acceleration is considered towards center (Centripetal)? what about centrifugal?
 
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  • #2
Hi Athi Sankar and welcome to PF.

If you spin a ball in a circle at the end of a string, the tension that changes the direction of the ball is directed towards the center. That is centripetal. The pull that you fell in your hand by the string is the reaction counterpart of the first force and is directed away from the the center. That is centrifugal. In short, what you call it depends on your frame of reference.

On Edit: If you want to make an object go around a circle at constant speed, the acceleration has to be perpendicular to the velocity. If the acceleration vector has a component in the same direction as the velocity, the speed will increase; if the acceleration vector has a component in the opposite direction as the velocity, the speed will decrease. The only way to keep the speed constant is to have no component along the velocity. To convince yourself that the acceleration is towards the center, put an "X" on a piece of paper and draw a circle around it. Note that you change the direction of the pencil always towards the "X" where you want the center of the circle to be.
 
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  • #3
The rate of change of velocity with respect to time is known as acceleration.A change in the velocity requires, A change in the magnitude (or) A change in the direction (or) A change in both magnitude and direction. As per the formula tangential Acceleration is the product of radius and angular acceleration (how fast the direction is changing). If the rate of change of direction (uniform angular velocity) is constant then Tangential Acceleration is Zero.But Radial Acceleration exists (Since it is change in magnitude not the direction). how do we take the direction of radial acceleration
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  • #4
Athi Sankar said:
how do we take the direction of radial acceleration
The direction of radial acceleration is radial, per defintion.
 
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  • #5
Athi Sankar said:
As per the formula tangential Acceleration is the product of radius and angular acceleration (how fast the direction is changing).
Correction: Tangential acceleration is how fast the speed in the direction of motion is changing, not how fast the direction is changing. If the tangential acceleration is zero, and the centripetal acceleration is not, the object will go around in a circle at constant speed; its direction of motion will change continuously. The centripetal acceleration is the one that changes the direction and is perpendicular to the tangential direction which makes it along the radial direction. Remember that the tangent to a circle at a given point is perpendicular to the radius from the center to that point.
 
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  • #6
kuruman said:
Correction: Tangential acceleration is how fast the speed in the direction of motion is changing, not how fast the direction is changing.
And this part is also backwards:
Athi Sankar said:
Radial Acceleration exists (Since it is change in magnitude not the direction).
Could it be that the OP confuses direction and magnitude?
 
  • #7
Velocity has the two characters (Magnitude and Direction). Rate of change of anyone or both character is the acceleration. In tangential Acceleration which one is changing? and also in Radial Acceleration (Rate of change) ? Thanks
 
  • #8
Athi Sankar said:
Velocity has the two characters (Magnitude and Direction). Rate of change of anyone or both character is the acceleration. In tangential Acceleration which one is changing? and also in Radial Acceleration (Rate of change) ? Thanks
Tangential acceleration changes the magnitude (which is the speed) but not the direction of the velocity vector.
Radial acceleration changes the direction but not the magnitude of the velocity vector.
The radial direction is always perpendicular to the tangential direction and at right angles to the direction of motion.
The tangential direction is always at right angles o the radial direction and parallel to the direction of motion.

But you do have to remember that the tangential and radial directions are continuously changing as the object moves along its circular path. Thus the statements above are always true, but they are talking about different directions at different times; an acceleration that is tangential at one moment won;t be quire exactly tangential a moment later. In the case of the rock on a string, the force is always radial, because the string is moving along with the rock in such a way that it is always lined up with the radial direction.
 
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  • #9
Doing this part first, since it hasn't been commented on yet.
Athi Sankar said:
what about centrifugal?
Assumr a puck on a frictionless surface attached to a string that goes through a hole in the surface and that the puck is traveling in a circular motion. There is a Newton third law pair of forces between the puck and string, the string exerts a centripetal force on the puck, and the puck exerts a centrifugal reactive force (a reaction to the centripetal acceleration) onto the string. (In a rotating frame, there is a fictitious centrifugal force, but I'm assuming this not what is being asked about).

Athi Sankar said:
Acceleration of a rotating link has two components,Tangential (change in the direction) Radial (change in the magnitude). Why the direction of Radial acceleration is considered towards center (Centripetal)?
The "radial" force is the force perpendicular to the velocity of an object, and points towards the instantaneous center of the curved path.

Going back to the puck on a frictionless surface and a string through a hole, note that if the string is pulled inwards, the puck follows a spiral like path, and that a component of the tension in the string is in the direction of travel of the puck, changing the speed of the puck as well as the direction. In this image, you can see that the force is not perpendicular to the path of the puck:

hole.jpg


If the string is winding or unwinding around a pole, then the tension will remain perpendicular to the puck, and the pucks speed will not change. The path is "involute of circle".

pole.jpg
 

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  • #10
Nugatory said:
The radial direction is always (...) at right angles to the direction of motion.
The tangential direction is always (...) parallel to the direction of motion.
rcgldr said:
The "radial" force is the force perpendicular to the velocity of an object, and points towards the instantaneous center of the curved path.
Is this common language? When I hear the words “radial” and “tangential” I automatically think of parallel and perpendicular to the position (from some origin), not to the velocity.
In other words, in my mind, “radial” and “tangential” are verbal references to a polar coordinate system.

Of course for circular motion, our interpretations are the same. But for general motion (like in @rcgldr’s spiral example) the meanings are different. In my view, I would say, “there is a radial component of velocity for non circular motion.”

Anyway, @Athi Sankar the general truth is that:

- The component of ##\frac{d\vec v}{dt}## parallel to ##\vec v## changes the magnitude of ##\vec v##

- The component of ##\frac{d\vec v}{dt}## perpendicular to ##\vec v## changes the direction of ##\vec v##

(##\vec v## does not even need to be a velocity, it could be any vector!)
 
  • #11
Hiero said:
Is this common language
Only in the case of circular motion, but that's what we're talking about in this thread.
 
  • #12
Nugatory said:
Only in the case of circular motion, but that's what we're talking about in this thread.
I had a feeling you were silently assuming circular motion, but @rcgldr explicitly says:
rcgldr said:
The "radial" force is the force perpendicular to the velocity of an object, and points towards the instantaneous center of the curved path.
With an example of spiral motion! Quite an unusual interpretation of the word radial.
 
  • #13
Hiero said:
I had a feeling you were silently assuming circular motion, but @rcgldr explicitly says:

With an example of spiral motion! Quite an unusual interpretation of the word radial.
At every point on the spiral trajectory the path is momentarily approximated by a circular arc (which will not be centered on the center of the spiral). One would often use such a curved tangent arc to determine how to split an acceleration into "radial" and "tangential" components.
 
  • #14
jbriggs444 said:
At every point on the spiral trajectory the path is momentarily approximated by a circular arc (which will not be centered on the center of the spiral). One would often use such a curved tangent arc to determine how to split an acceleration into "radial" and "tangential" components.
I’m familiar with the concept of curvature, but does anyone actually mean that by “radial”...?
I have only ever heard that basis referred to as “tangent, normal, (bi-normal)”

It just seems odd to call the normal “radial,” but if people do call it that, then that’s fine, I am not the decider of semantics.
 
  • #15
Hiero said:
It just seems odd to call the normal “radial,” but if people do call it that, then that’s fine, I am not the decider of semantics.
Perhaps this discussion can be put in perspective if one considers the expressions for velocity and acceleration in polar coordinates for motion in a plane. Starting with ##\vec r(t)=r~\hat r##, one obtains the general expressions$$\vec v(t)=\frac{d\vec r}{dt}=\dot r~\hat r+r\dot \theta~\hat \theta$$ $$\vec a(t)=\frac{d\vec v}{dt}=(\ddot r-r\dot \theta^2)~\hat r+(r\ddot \theta+2\dot r \dot\theta)~\hat \theta$$Common special cases,
1. Linear motion, ##\dot \theta = \ddot \theta = 0## and ##\vec a(t)=\ddot r \hat r##.
2. Circular motion ##\dot r =\ddot r =0## and ##\vec a(t)=-r\dot \theta^2~\hat r+r\ddot \theta~\hat \theta##.
In all cases, whatever multiplies unit vector ##\hat r## is the "radial" component and whatever multiplies ##\hat \theta## is the "tangential" component much like ##\vec A(x,y)=f(x,y)~\hat x+g(x,y)~\hat y## wherein whatever multiplies ##\hat x## is the "x-component" and whatever multiplies ##\hat y## is the "y-component".

However, we still need to hear from OP whether the original question has been answered.
 
  • #16
Hiero said:
I’m familiar with the concept of curvature, but does anyone actually mean that by “radial”...?
I have only ever heard that basis referred to as “tangent, normal, (bi-normal)” It just seems odd to call the normal “radial,” but if people do call it that, then that’s fine, I am not the decider of semantics.
There is radius of curvature, and radial acceleration could be considered as acceleration in the direction of radius of curvature. Radial acceleration is a synonym for centripetal acceleration. Links:

http://www.ux1.eiu.edu/~cfadd/1150/03Vct2D/accel.html

https://www.quora.com/What-are-the-...tripetal-acceleration-and-radial-acceleration
 
  • #17
rcgldr said:
Radial acceleration is a synonym for centripetal acceleration
I don't believe so. If you look at post #15, the radial component of the acceleration has two terms. The first term is the rate of change of the radial speed and the second term is the centripetal term. Example: A car is on a rotating platform and the driver pushes on the accelerator pedal along a radius of the platform. The rate at which the speedometer needle rises is a measure of ##\ddot r## and has nothing to do with the centripetal term ##-r \dot \theta^2##. They should not be both lumped together as "centripetal".
 
  • #18
kuruman said:
I don't believe so. If you look at post #15, the radial component of the acceleration has two terms. The first term is the rate of change of the radial speed and the second term is the centripetal term. Example: A car is on a rotating platform and the driver pushes on the accelerator pedal along a radius of the platform. The rate at which the speedometer needle rises is a measure of ##\ddot r## and has nothing to do with the centripetal term ##-r \dot \theta^2##. They should not be both lumped together as "centripetal".
It seems to me like merely a matter of what center you are using. The momentary center of curvature or the center identified in the coordinate system.
 
  • #19
rcgldr said:

kuruman said:
I don't believe so.
Take a look at the links I referenced, in the vector diagrams where acceleration is split into normal and tangential components, the normal component is also called centripetal or radial. I only posted links to those two articles as examples but you can find others doing a web search for "radial acceleration" .
 
  • #20
rcgldr said:
Take a look at the links I referenced, in the vector diagrams where acceleration is split into normal and tangential components, the normal component is also called centripetal or radial. I only posted links to those two articles as examples but you can find others doing a web search for "radial acceleration" .
I did take a look at the links and I do not disagree with what is said. However, in those links the motion is limited to circular, namely to cases where ##\dot r=0##. My point in #15 and #17 is that there may be another radial term when the motion is not restricted to a circle and that therefore the centripetal term is only part of the radial acceleration, not the whole story. As @jbriggs444 pointed out in #18, this assumes that a fixed center is used to measure angles and radial distances. If one insists that the only radial component is centripetal, then one has to approximate an arbitrary curved path as a succession of arcs ##ds## each being part of a circle with its own radius of curvature and center. However, I am not sure that this description is more advantageous for formulating solutions to problems than the fixed center description.

We have not heard from OP for about 48 hrs. I will suspend posting on this thread until we do.
 
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  • #21
kuruman said:
I did take a look at the links and I do not disagree with what is said. However, in those links the motion is limited to circular, namely to cases where ##\dot r=0##.
The third diagram from the first link (ux1.eiu.edu) shows a changing path, with total acceleration separated into separate components, ##a = a_r + a_t## or ##a = a_c + a_t## where ##a_r## is radial acceleration and a synonym for ##a_c## which is centripetal acceleration. The path would not have to be a circle, it could be an ellipse, spiral, parabola, hyperbola. Consider a car traveling along a winding road with increasing and decreasing radius turns. The car can be braking or accelerating as well as turning while following the curves in the road, with both tangential and radial acceleration.
 
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  • #22
kuruman said:
We have not heard from OP for about 48 hrs
I was preparing for exams. So I could not reply for a while
rcgldr said:
The "radial" force is the force perpendicular to the velocity of an object, and points towards the instantaneous center of the curved path
This Explanation was sufficient from my subject point of view and I thank all for taking effort to clear my doubts.
 

1. Why does radial acceleration act toward the center?

The direction of radial acceleration is always towards the center because it is caused by the centripetal force, which is directed towards the center of the circular motion. This force is necessary to keep an object moving in a circular path.

2. How is radial acceleration related to centripetal acceleration?

Radial acceleration and centripetal acceleration are essentially the same thing. Radial acceleration is the component of centripetal acceleration that is directed towards the center of the circular motion.

3. Does radial acceleration change as an object moves in a circular path?

Yes, the magnitude of radial acceleration changes as an object moves in a circular path. This is because the velocity of the object is constantly changing, and since radial acceleration is dependent on velocity, it also changes.

4. What factors affect the magnitude of radial acceleration?

The magnitude of radial acceleration is affected by the speed of the object, the radius of the circular path, and the mass of the object. A higher speed or smaller radius will result in a larger radial acceleration, while a larger mass will result in a smaller radial acceleration.

5. Can radial acceleration be negative?

Yes, radial acceleration can be negative. This occurs when the object is slowing down and the direction of the acceleration is opposite to the direction of motion. In this case, the centripetal force is acting in the opposite direction, causing the object to decelerate.

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