# Question about circular motion and acceleration

parshyaa
In circular motion
1) V = rw and ##\vec V## = r ω##\vec e_{tan}##
2) a = rα and ##\vec a## = -##\frac{v^2}{r}####\vec e_{rad}## + rα##\vec e_{tan}##
Where ##\vec e_{tan}## is the unit vector along the tangent in increasing direction of θ
From 1) we see that rω is the magnitude of velocity of particle executing circular motion and its direction is along tangent
But in 2) we see that magnitude of acceleration is rα but this is not the magnitude of total acceleration
How could you explain that rα is not the magnitude of total α

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But in 2) we see that magnitude of acceleration is rα
This is not correct. It only describes the tangential acceleration.

parshyaa
This is not correct. It only describes the tangemtial acceleration.
I know that, this is not the magnitude of acceleration, it will be root of the sum of the square of components in both tangential and radical direction, but in v=rω its the total magnitude of the velocity and in a= rα its not the magnitude of total acceleration, its just the magnitude along the tangent, this is what not satisfying me, in the previous case total magnitude is same as v= rw, but this is not the case with accelaration, why is this so

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Take a general velocity ##\vec v##. Its magnitude, i.e., the speed, ##v## satisfies ##v^2 = \vec v^2## which means that the time derivative ##\dot v## of the speed is given by the time derivative of this expression as
$$\frac{dv^2}{dt} = 2v\dot v = \frac{d\vec v^2}{dt} = 2\vec v \cdot \frac{d\vec v}{dt} = 2\vec v \cdot \vec a. \quad \Longrightarrow \quad \dot v = \frac{\vec v \cdot \vec a}{v^2}.$$
Hence, only the acceleration in the direction parallel to ##\vec v## matters for the change in the speed. Any acceleration orthogonal to the velocity will only change the direction, but not the speed.

parshyaa
parshyaa
Take a general velocity ##\vec v##. Its magnitude, i.e., the speed, ##v## satisfies ##v^2 = \vec v^2## which means that the time derivative ##\dot v## of the speed is given by the time derivative of this expression as
$$\frac{dv^2}{dt} = 2v\dot v = \frac{d\vec v^2}{dt} = 2\vec v \cdot \frac{d\vec v}{dt} = 2\vec v \cdot \vec a. \quad \Longrightarrow \quad \dot v = \frac{\vec v \cdot \vec a}{v^2}.$$
Hence, only the acceleration in the direction parallel to ##\vec v## matters for the change in the speed. Any acceleration orthogonal to the velocity will only change the direction, but not the speed.
Wonderfull explanation, just replace v^2 by v
Thank you so much, for this explanation