# Tangential and Centripetal Acceleration in Circular Motion

• andyrk
In summary, the conversation discusses the absence of tangential acceleration in uniform circular motion and the presence of both tangential and centripetal acceleration in non-uniform circular motion. The Work-Energy Theorem is used to explain why the angle between acceleration and displacement is always 90 degrees in uniform circular motion, making the acceleration purely centripetal. It is also mentioned that the method provided does not explain how centripetal acceleration changes the direction of the particle, but it can be shown mathematically using equations of motion. The conversation also touches on the distinction between tangential and centripetal acceleration and their effects on the speed and direction of the particle. Finally, it is discussed that centripetal acceleration acts as a "perpetually
andyrk said:
But I don't think it won't have any explanation.
No explanation beyond the math already posted

andyrk said:
I never avoided calculus.
But you seem to be avoiding what it is telling you. The point of describing Physics (your sporting prfeormance and your Bank Balance) with maths is that Maths describes the relationships between Physical (and other) quantities very well and it predicts outcomes. What more is required? You seem to want to have a verbal narrative running in parallel with what the Maths is telling you. Why should this be necessary?
Verbal speech is just not adequate for dealing with the more sophisticated relationships that constitute modern Science.

Velocity is a vector, so it has a magnitude and a direction. If something changes the vector but does not change the magnitude then it must change the direction.

It seems that you already understand how the component parallel to the velocity is the component that changes the magnitude. Since the acceleration component perpendicular to the velocity changes the velocity but does not change its magnitude then it must change its direction.

You can also show this as follows. Without loss of generality, we consider a coordinate system where the velocity vector and the acceleration vector are both in the x-y plane and the velocity vector is aligned with the positive x axis. The magnitude of the velocity is:
##s=\sqrt{v_x^2 + v_y^2 + v_z^2} = v_x ##
##\dot{s}= (v_x \dot{v_x} +v_y \dot{v_y} +v_z \dot{v_z} )/(\sqrt{v_x^2+v_y^2+v_z^2}) = \dot{v_x} = a_x ##

The direction of the velocity is:
##\theta = \arctan(v_y/v_x) ##
##\dot{\theta} = (v_x \dot{v_y} - v_y \dot{v_x})/(v_x^2+v_y^2) = \dot{v_y}/v_x = a_y/v_x##

So the change in speed is due to the acceleration in the x-direction, which is parallel to the velocity, and the change in direction is due to the acceleration in the y-direction, which is perpendicular to the velocity.

andyrk said:
Yeah, I know that. This is maths and not physics. But you still didn't explain that how ar changes the direction of the particle to keep it moving in a circle. My question is so simple, yet no one is able to answer it. I wonder why.

You are aware that it is the centripetal force that causes the change in direction, which is what centripetal acceleration is, right? The particle is under a net force which changes its velocity, and that change in velocity is acceleration. The acceleration doesn't cause the particle to move in a circle, it is the result of the particle being forced to move in a circular path by a force.

In case of a uniform circular motion, let us add two constant velocities in a chosen x and y directions. In other words, we shift the frame of reference to another inertial frame of reference.The acceleration components would obviously 0, since the velocity components of new frame of reference are constants. Thus the uniform circular motion shall remain a uniform circular motion under transformation into inertial frames of reference and there would be no chance in radial and tangential velocities under such transformation.

@andyrk I don't want to hijack your thread, it's interesting. But I also have a simple question on centripetal force or acceleration.
I decided not to start a new thread for such small info.

Can anyone prove mathematically F=(mv2)/R ? or acceleration a=v2/R ?
All proofs I have seen so far in books etc are Graphical, using vector addition method. Again prove by using simple Mathematics not Graphs.
Thanks.

Neandethal00 said:
@andyrkCan anyone prove mathematically F=(mv2)/R ? or acceleration a=v2/R ?
All proofs I have seen so far in books etc are Graphical, using vector addition method. Again prove by using simple Mathematics not Graphs.
Thanks.
See post #20 with $\ddot \theta = 0$.

DrGreg said:
See post #20 with $\ddot \theta = 0$.
Yes, it does show a=v2/R, Thanks.
I wonder why no book uses such simple derivation instead of geometric hoopla?
Doesn't it also tell us acceleration of a rotating mass tied to string length R is the same as acceleration of
a rotating rod of length R?

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