1
$\begingroup$

Let us say a particle is moving in a circle with Uniform circular motion .

Then , if I take an equation like this .

$\theta$ = S(Arc length or Total distance covered ) / Radius of circle *( time taken to cover this distance ) = $\delta$ v ( final - initial velocity ) / v final or initial *( Time taken )

= Velocity of particle / Radius of circle = Centripetal acceleration/ Radius of circle .

So , I am not getting is why do we get centripetal acceleration here.

Why not tangential or total acceleration?

Improve this question
$\endgroup$
4
  • $\begingroup$ Have you researched the words centripetal and centrifugal? $\endgroup$ – Solar Mike Feb 4 at 20:45
  • $\begingroup$ @SolarMike Not centrifugal. It is related to non inertial frame also . A bit confusing then. I am new to circular motion . But I do have an analogy of my own that I thought . You can check it in the comments of tiger guy $\endgroup$ – Srijan M.T Feb 4 at 20:55
  • $\begingroup$ Your equations don't make sense, and the answer is wrong (the acceleration is not $v/r$, $v/r$ doesn't even have the right dimensions for an acceleration.) You can't try to apply SUVAT equations to something moving along a curve - except in the special case of a projectile where the horizontal and vertical motions are independent of each other. $\endgroup$ – alephzero Feb 4 at 23:06
  • $\begingroup$ @alephzero This was done by my teacher . $\endgroup$ – Srijan M.T Feb 5 at 8:28
1
$\begingroup$

It's there to make the laws of motion work.

An object in stable circular motion needs a force towards the center of the circle to keep it in circular motion.

f=ma

Pretty simply the force required to keep it circling results in "acceleration," which here is defined as centripetal acceleration. Note that for an object in orbit, that force is basically gravity, which fits the equation. This is not the traditional concept of acceleration meaning an object speeding up over time. Think of it more as the object's velocity vector changing over time. If your spinning body had a force on it making it speed up, we could also have angular acceleration. I'm not familiar with terminology around tangential acceleration or "total" acceleration.

Improve this answer
$\endgroup$
1
  • $\begingroup$ I am not getting exactly what you mean to say . What I think from your answer is that centripetal acceleration is written there since tangential acceleration is 0 in Uniform circular motion. Therefore , we can say total acceleration or centripetal acceleration. Both have same values . Is this right ? $\endgroup$ – Srijan M.T Feb 4 at 20:54
1
$\begingroup$

I found this definition quite interesting, you can comment on it.

"Centripetal acceleration is defined as the property of the motion of an object, traversing a circular path. Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration."

enter image description here

Better picture.

enter image description here

Improve this answer
$\endgroup$
0
$\begingroup$

Your nomenclature and variable name choices are a horror. I'll try and redo your stuff in the canonical way

Rotational velocity $\displaystyle = \omega = \frac{s}{r\cdot t} = \frac{\theta}{t} \ $seconds$^{-1} $

Acceleration has the units of meters per second$\,^2$, the formula for centripetal acceleration is $\omega^2\cdot r$

In steady rotation this is also the total acceleration.

In polar coordinates, the unit vectors are radial and tangential. The centripetal acceleration is radial by definition, and the tangential acceleration is zero.

In cartesian coordinates, the $x$ and $y$ accelerations vary depending on $\theta$, Where $\theta=0$ corresponds to the $\raise{.1 em}{\tiny{+}}\normalsize{x}\ $axis and runs CCW to the $\raise{.1 em}{\tiny{+}}\normalsize{y}\ $axis

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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