# Can centripetal acceleration be called centripetal? jerk?

1. Jun 18, 2011

### Alpharup

Well I am now confused about centripetal acceleration. I have heard sometime ago that the change of acceleration is jerk. Let a body in centripetal acceleration change its velocity from 'v' to 'v1' in time 't'. The formula for centripetal acceleration is v^2/r. Let this acceleration be 'a'. Can the acceleration which originates from the point v1 be considered as v1^2/r? If it can be considered let this acceleration be 'a1'. Since the acceleration changes from 'a' to 'a1' in time 't', can the rate be called jerk? is my arguement correct?

2. Jun 18, 2011

### Staff: Mentor

I don't quite understand what you're saying here. In any case, since jerk is the rate of change of acceleration (a vector), even uniform circular motion displays jerk.

3. Jun 18, 2011

### Andrew Mason

As you point out, the third time derivative of position for an object experiencing uniform circular motion is non-zero. Whether this should be called "jerk" is a matter of terminology. I don't know how that term is supposed to be used - it is rarely used by physicists. But there is nothing "jerky" about uniform circular motion.

If the term "jerk" was intended to refer to the inertial effects within a body arising from a change in applied force, it would be confusing to apply it to uniform circular motion, which gives rise to no such effects.

AM

4. Jun 19, 2011

### Staff: Mentor

I think you're right. I've never seen it applied in that fashion. (And certainly from a rotating frame of reference there'd be no changing acceleration.) I've only seen jerk used to describe linear motion. (By engineers, not physicists.)

5. Jun 21, 2011

### timthereaper

Centripetal acceleration in uniform circular motion only changes in direction, not in magnitude. From my experience in engineering, "jerk" is the change in acceleration per unit time, but it usually only applies to the change in magnitude, not the change in direction. If the object were speeding up or slowing down, you could describe that as "jerk". However, I've only heard the term used rarely and only where sudden changes in acceleration become important, like in cam design.

6. Sep 23, 2011

### Alpharup

wellve i have found an expression for this jerk. please correct me if iam wrong. for a body to revolve in uniform circular motion, you need a force which is called centripetal force.if this centripetal force is constant the body revolves with constant speed . inorder to induce tangential acceleration to the body in circular motion, you need to increase the force at a constant rate. let us assume that the body starts from rest and moves with constant speed for a time 'x'. here the initial centripetal force is constant. let the angular velocity of the particle be 'w' and linear velocity be 'v' for this constant centripetal force. after this time x, you increase your force so as to induce tangential acceleration 'A'. the resultant of the centripetal and tangential acceleration can be found out. the body also experiences a jerk[J] which is directed towards the centre. the expression is
J=2wA

the proof is very simple as i had said in my first post ''Let a body in centripetal acceleration change its velocity from 'v' to 'v1' in time 't'. The formula for centripetal acceleration is v^2/r. Let this acceleration be 'a'. Can the acceleration which originates from the point v1 be considered as v1^2/r? If it can be considered let this acceleration be 'a1'. Since the acceleration changes from 'a' to 'a1' in time 't', can the rate be called jerk?'' let us use these variables which i used earlier but with some change. let the change in velocity of the body be very small ie.. v1~=v. let the time for of change also be very small ie.. t tends to '0'. i had used a little calculus to prove it.
now J=a1-a/t.
by simplification and by substituting the values of acceleration
we get
J= v1^2 - v^2 / rt
= [v1+v]*[v1-v]/rt {using algebra equation}
= [v1+v]*A/r {since the time to change is small, rate of change of velocity gives tangential acceleration}
= 2vA/r {since velocity change is small, we can substitute v1=v}
= 2wA {since v=rw or linear velocity = [angular velocity]*[radius]}

this even satisfies cross product of vectors where direction of jerk can be found that it is towards the centre
I think I have proved.

7. Sep 23, 2011

### A.T.

You are apparently assuming that the body in circular motion is also spinning at the same rate, so the centripetal force vector is constant in the reference frame of the body. But this is not always the case. If you are in a cabin that goes in circles, but keeps a fixed orientation, you will be "jerked around" a lot.

8. Sep 23, 2011

### Andrew Mason

You are quite right. That is part of the principle behind the Tilt-o-Whirl and other amusement rides. In that case, however, you would not be travelling in uniform circular motion.

AM

Last edited: Sep 23, 2011
9. Sep 23, 2011

### A.T.

Why not? Every part of of the cabin travels in uniform circular motion, just around different centers. I would be "jerked around" in the cabin, just like in a car that truns left - right - left. The word "jerk" makes perfect sense here.

10. Sep 23, 2011

### Alpharup

well is my expresion, right?

11. Sep 25, 2011

### Andrew Mason

It does indeed make sense. My only point is that it is not in uniform circular motion. The OP asked whether uniform circular motion (ie where $\vec{a} = -\hat r v^2/r$) could be termed "jerk".

The centre of mass of the cabin moves in uniform circular motion. But if you plot the motion of a person in one of the outer corners you will see that the radial displacement from the centre of rotation is constantly changing in both magnitude and direction and the angular speed varies as well.

AM