A body that moves in a circular motion (of radius *r*) at constant speed (*v*) is always being accelerated. The acceleration is at right angles to the direction of motion (towards the center of the circle) and of magnitude *v*^{2} / *r*.

The direction of acceleration is deduced by symmetry arguments. If the acceleration pointed out of the plane of the circle, then the body would leave the plane of the circle; it doesn’t, so it isn’t. If the acceleration pointed in any direction other than perpendicular (left or right) then the body would speed up or slow down. It doesn’t.

Now for the magnitude. Consider the distance traveled by the body over a small time increment Δ*t*:

We can calculate the arc length *s* as both the distance traveled (distance = rate * time = *v* Δ*t*) and using the definition of a radian (arc = radius * angle in radians = *r* Δ*θ*:)

The angular velocity of the object is thus *v* / *r* (in radians per unit of time.)

The right half of the diagram is formed by putting the tails of the two *v* vectors together. Note that Δ*θ* is the same in both diagrams.

Note the passing from sin to cos is via l’Hôpital’s rule.

*QED*.

But it doesn't give you direction, whereas acceleration is a vector quantity

Jack I believe the direction is given by v, because it is also a vector. This equation can be rewritten vector component wise.

plz make this more easier .i does not understand this .

Express in a easy way. With out using vectors notation.

nothing special

where did you get sin(dtheta/2)?

We need linear velocity and the radius for solving… And we know that sin angle is equal to opp side upon hypotenuse…

@Daniel

I got sin(Δθ/2) = (Δv/2)/v as follows.

Look at the right-hand side of the figure diagram; it's a triangle with two red arrows and a green arrow. There's a black line from the place where the two red arrows meet, to the midpoint of the green arrow.

There's a right triangle made up of the top red arrow, the top half of the green arrow, and the black line.

The hypotenuse of this right triangle is v. The side made up by half of the green arrow is Δv/2. The angle opposite the green side is Δθ/2.

sin = opposite/hypotenuse, so sin(Δθ/2) = (Δv/2)/v.

thnqu very Much…….Now I UNDERSTOOD this derivation

Oh I see it now, thank you for the explanation [MSFT].

Can't get it,s0 tense ab0ut this and just g0nna cry :-(

thats nice

mr msft can you please explain 5line of the second derivation

sin (Δθ/2) and Δθ/2 both tend to 0 as Δθ tends to 0, so we have the indeterminate form 0/0.

We can use l'Hôpital's rule; take the derivative of the top and the bottom with respect to Δθ.

Top: (sin (Δθ/2))' = (cos (Δθ/2))((Δθ/2)') = (cos (Δθ/2))(1/2). This tends to 1/2.

Bottom: (Δθ/2)' = 1/2. This tends to 1/2.

thanks

it could be made much more easier without using trigonometry in it….

from sin delta theta/2 =deltaV/2 hw did deltaV become 2V sindeltatheta/2

from sin delta theta/2 =deltaV/2 hw did deltaV become 2V sindeltatheta/2

from sin delta theta/2 =deltaV/2 hw did deltaV become 2V sindeltatheta/2

i gets the lesson of my teacher

very good but i can't understand anything in the formula!

Nicely done thank you Maurits , this is the simplest way to derive the centripetal acceleration

bakwaas hai

Oh finaly I got it .. thx a ton

There's a far simpler derivation than this. Draw the circle with the two velocity vectors separated by a tiny angle. Draw a vector diagram that shows that dv = vdq. Divide both sides by dt, then a = vw.=v2/r.

It's a beautiful proof :)

wow

complicated ….

i have not understand the third stage????

Mr Maurits u have explained it super …..love ur w

ay man its very nice no use of those shitty components simple……. thnx man

Mr Maurits u have explained it super love ur way simple and easy the best part u have done is that u have avoided the use of those components that confuse..can I use this method all this I hopitals rule for an 11 the class exam?

Make it a bit simpler,please!

For the limit as delta theta [D0] approaches O of v*sin(D0/2)/D0/2, can't the sine component be thought of as its infinite series too, with (x – x^3/3! + x^5/5! +…)/x = 1 – x^2/3! + x^4/5! +…, so as x approaches O, or D0 does so, the series approaches 1; making the initial equation = v*1 = v

I came here trying to understand the underlying mathematics, it's just when someone uses someone else's theorems or whatever in their derivation, I then spend forever trying to understand the utilised theorems and end up on an infinite search for understanding. The infinite series method just helped me jump one of those hurdles — and I'm hoping it helps someone else like minded :P

Yes, that is sensible.

Nice style

Easiest one

coud you please explain me what is i'hopital's rule?

Its soo difficult to understand…..

Not understanding

Two bodies of masses 10kg and 5kg movingvin concentric orbits of radii R and r such that their periods are same. Then the ratio between their centripetal accelerations is:

(a) R/r

(b)r/R

(c)R^2/r^2

(d)r^2/R^2

plz help me give me the solution of this question

fabulous explanation

Thank you for posting a simple and easy to understand mathematical derivation of the equations. I had to google this and the first 4 links weren’t helpful. You were the 5th link. Also thanks for posting the note Re: L’Hopital’s rule. I did not get that step until I saw your note. PERFECT!

Hey bro is it easy??? Then u make it more easy for us

Hard to understand. Very lengthy derivation. Please shorten and post.

Hard to understand. Very lengthy derivation. Please shorten and post..

Its very difficult plzz try to explain it in a easiest way😦

Can’t undrstand plzz try to explain in easist way😦

It is too much lengthy and difficult to understand sir