0
$\begingroup$

Say we have a hinged rod resting horizontally. If we apply a force at the unhinged end of the rod there will be a tangential acceleration of the end point of the rod.

Now, does this point also already have a centripetal acceleration? The point isn't moving in a circle at this instant but it will be moving in a circle the instant after.

Note I say "at this instant" in the calculus sense of the word.

Why/ why not? How to approach such questions?

Improve this question
$\endgroup$
3
  • $\begingroup$ Before this question can be answered you will have to explain how you would go from "no force" to "a (finite) force" instantaneously. If you can do this then you can answer your own question. $\endgroup$
    – Farcher
    Commented Dec 17, 2018 at 12:13
  • $\begingroup$ I guess what you're saying is that anything "instantaneously" isn't possible? But how can the situation be the same for a point that is already moving in a circle and a point that is resting? $\endgroup$
    – delivosa
    Commented Dec 17, 2018 at 12:17
  • 1
    $\begingroup$ If you use a mathematical model you can apply a force instantaneously and then the rod has an acceleration but zero velocity at the instant that you apply the force. $\endgroup$
    – Farcher
    Commented Dec 17, 2018 at 12:41

2 Answers 2

Reset to default
1
$\begingroup$

How could it not move in a circle. You have a rigid constraint that mandates that the only motion any point on the rod can make is circular motion. As soon as the velocity is non-zero the centripetal acceleration will be non-zero. If it starts at rest then at the instant ac = 0. Once omega is non-zero ac is non-zero.

Improve this answer
$\endgroup$
4
  • $\begingroup$ But at the instant that the force applies it isn't moving in a circle, it is at rest. $\endgroup$
    – delivosa
    Commented Dec 17, 2018 at 12:05
  • $\begingroup$ How can the situation of a resting point and a point that is already moving in a circle be the same? $\endgroup$
    – delivosa
    Commented Dec 17, 2018 at 12:13
  • $\begingroup$ You seem to be creating a scenario where something moves and dosen't move at the same time. See the other answer provided by melp. At the instant you apply a force you create an instantaneous rate of change for v, a change of state. Motion is continuous and you cannot separate the "instant" from neighboring instants. $\endgroup$
    – user196418
    Commented Dec 17, 2018 at 12:34
  • $\begingroup$ At the instant omega = 0, centripetal acc = 0. From my point of view the constraint of the hinge makes it all together illegal to not move in a circle (once motion is set up). $\endgroup$
    – user196418
    Commented Dec 17, 2018 at 12:35
1
$\begingroup$

From the instant the force is applied , we should use two formulas to help us understand what physics is at play. So we want to know how force is related to acceleration, and specifically centripital acceleration. Luckily , physics is described by mathematics , which have very specific, and strict rules, and definitions.

Let’s understand general force first, Force = massxacceleration re-arranging , acceleration = Force / mass Providing force , is non zero, we will have acceleration (tangential) over time , so let’s break down the centripital formula in the same way,

Centripetal acceleration is the rate of change of tangential velocity.

Then if we are now looking at “a change” in the tangential velocity vector , at time = 0 , there is no change , as we need two points in time to find a difference.

velocity is distance over time = d/ t

so we need Fc = mac = m v^2/ r = m*angularFrequency^2*r

Hence , the question you need to ask is,

Has the tangential velocity changed over time ? Has the direction of travel changed?

When we consider calculus , and taking a derivative , what we are doing is finding the instantaneous rate of change. So providing you have non-zero figures , you will have non-zero centrifugal acceleration.

Hopes this helps you get started thinking about how to approach your problem

https://theory.uwinnipeg.ca/physics/circ/node6.html

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.