Derivation of Centripetal Acceleration (without calculus)

[@Omiknowest]

The best ever you are. All these KZbinrs got me going nuts all along. Please remain blessed

[@DotPhysics]

thank you very much

[@manujamethsith]

Wish I saw his explanation before cuz I've been unable to understand this in school for long time .Superb explanation and I will show this to my mates and every understand this in one go cuz explanation super easy even my dog can understand this .THANKYOU SOOO MUCH FOR SAVING MY LIFE🙂🙂🙂

[@jameskloberdance9338]

Fun derivation! I love to show this to my geometry students after they learn about similar triangles.

[@Dameng123]

your explanation was very easy to understand.

[@lipa9121]

i can’t understand according to which geometrical proof Δθ of the first triangle is equal to Δθ of the other one

[@arandompersononyoutube3739]

Well that's because Angle between two lines is same as angle between their perpendiculars. 4:17. Both v1 and v2 are perpendicular to the radius r. So the angle between the two radii (∆θ) is same as the angle between their perpendiculars v1 and v2 (Δθ).

[@josuelima5033]

to sum two vectors you need to connect their origins. Radius vectors is always perpendicular to tangent vectors like velocities (r1 and v1; r2 and v2). r1 and r2 are also connected by their origins, so you can visualize that as if this radius set rotates 90º it will get the velocities set direction, and this doesnt change the angle between them

[@pheonixxt221]

Very useful class, sir.. ❤ Love it

[@DannyLanghamer]

Very well explained, definitely will use to help me study for A level Physics

[@warunilokuge8472]

Thanks so much , I wasn't sure as to whether the direction of the change in velocity was towards the centre of the circle or not and thanks for clarifying it.

[@user-hd3hy2gg9f]

4:18 if anyone wants the proof of the angle between two radius is the angel between the two tangent lines, u can ask me.

[@urvikhadse]

Hey, I'm asking you.... please tell me the proof

[@ilofek]

bro i would love to have one

[@mohamedgamal-fl1lb]

but unclear why are triangles are similar ? plus unclear delta V takes this direction ?

[@ferhatnusreturuc1809]

very good explaination, thank you

[@Chris-rt3vk]

4:42 You lost me there. How do you get that expression for ∆s?

[@aswathynnn]

We have the formula theta=arc length/radius Here arc length is delta ,by multiplying delta s=R delta theta

[@mrbigcat9]

If the magnitude of the velocities are the same why is there a delta V?

[@DotPhysics]

It's because velocity is a vector. So, a change in direction IS a delta v

[@EstherShedrach-ut1oc]

Show that the magnitude of the centripetal acceleration for a uniform circular motion is given as Ac=V2 =W2r r

[@badermohsin316]

Awesome video, although i didn't understand how velocity pints to the center

[@lkuty]

acceleration points to the center, not velocity which is orthogonal to position vector (tangential to the circle)

[@aswathynnn]

But change in velocity delta v points to center right

[@Dameng123]

thanks a lot

[@muhamadym3403]

5:57 How did u get ac?

[@DotPhysics]

the acceleration is change in v divided by change in t. I just sub in my expression for delta v (in terms of theta) and simplify

[@User-jr7vf]

Why is it not a complete derivation?

[@DotPhysics]

This just gives the magnitude and not the direction (although I make an argument for that). I think the best derivation is to look at the r-hat component of acceleration in polar coordinates - here's my video on that kzbin.info/www/bejne/e4qzqHeMdrOiocU

[@User-jr7vf]

@@DotPhysics oh I see, thanks!

[@fizixx]

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