Can somebody show me tha derivation of this formula?

Question

a=αr ----->this formula relates the linear acceleration and the angular acceleration


Our teacher just gave us this formula and didn't show us how it was derived.

Oooopsss..
I forgot:
a=linear acceleration
α=angular acceleration
r=radius

Answer 1

Your teacher would be unable to derive such fomula, because, simply put, this formula is wrong.

Correct formula is
a = -R ω² + α x R, where R denotes vector (x,y)
and its derivation is:
v = ω x R
a = dv/dt = dω/dt x R + ω x dR/dt
a = α x R + ω x v
a = α x R + ω * (ω x R)
and when R is perpendicular to ω
a = α x R - R ω²


People, you have no right to differentiate speed, you must differentiate VELOCITY. Otherwise you lose centripetal acceleration.

Answer 2

If you draw a small angle (theta) you can see that this translates into a small linear distance D= thetaX R, the radius relating the angle and distance.

The linear speed is S= dD/dt = d (theta)/dt X R= omega X R, where omega = d(theta)/det, the angular velocity.

The linear acceleration is a = dS/dt = R X d(omega)/dt = RXw
where w = angular acceleration.

Answer 3

Right, it's just from the definition of angle: s=(theta)*r. Take two time derivatives to get a = (alpha)*r.

Answer 4

From trig:

Remember unit circle ?

Arclength = (Angle) Radius

s=@r