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Acceleration is the rate of change of velocity. In the uniform circular motion the acceleration is produced due to change of direction of the velocity(the magnitude remains same). The direction is changing at a uniform rate depending upon the speed of the body. Then why is it said that acceleration is variable? What is that thing that is variable?

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    $\begingroup$ Didn't you just explain how the acceleration vector is changing direction constantly? Something changing is variable. The acceleration magnitude is constant and non variable. Similarly, the velocity vector is variable, while the velocity magnitude (the speed) is constant. $\endgroup$ – Steeven Feb 1 at 10:18
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Acceleration is a vector quantity. It has both magnitude and direction. Just as you said in the question, the direction of acceleration is changing in uniform circular motion. Therefore acceleration is variable, as a vector quantity is said to be variable if either its magnitude or its direction is changing.

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  • $\begingroup$ Then what about uniform acceleration? I said that everything is changing uniformly then why is acceleration variable, not uniform? $\endgroup$ – Kushagra Shukla Feb 1 at 10:51
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    $\begingroup$ @KushagraShukla - "Uniform acceleration" that doesn't change direction means linear motion, not circular. In circular motion, only the magnitude of the acceleration does not change. The direction of acceleration always points towards the circle center, which is a different direction at each point on the circle. $\endgroup$ – Paul Sinclair Feb 1 at 17:58
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The acceleration of a body travelling at uniform speed, $v$ (constant speed would be clearer) in a circle of radius $r$ is given by$$\vec a=-\frac{v^2}{r}\hat r$$ in which $\hat r$ is the unit radial vector joining the circle centre to the body at the time when its acceleration is $\vec a.$ Because $\hat r$ is continuously changing (I think this is clearer than 'variable', which, to me, means 'might change') so is the acceleration. The acceleration is changing at a rate$$\frac{d\vec a}{dt}=-\frac{d}{dt}\frac{v^2}{r}\hat{r}=-\frac{d}{dt}\frac{v^2}{r^2}\vec{r}=-\frac{v^2}{r^2}\frac{d\vec r}{dt}=-\frac{v^2}{r^2}\vec{v}=-\frac{v^3}{r^2}\hat v$$ $\hat v$ is the unit vector in the direction of the velocity, so tangential to the circle. Because $\hat v$ is continuously changing, so is the rate of change of acceleration! And so on.

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This is a common point of confusion for students. As other answers have stated, velocity, being a vector quantity, has magnitude and direction. We know that an object's velocity will normally be in a straight line, and the only way to change the velocity is to accelerate or decelerate the object (i.e. we must use force). To make something go in a circle, we have to constantly change the direction of the force, so we have to constantly change the direction of the acceleration. Therefore, acceleration is constantly changing.

Simply because we know how something changes, and that it changes continuously and uniformly, does not change the fact that it is changing. Similarly, a quantity which is changing continuously is, by definition, variable. Thus the acceleration of a particle in 2D space is changing continuously, and is also variable, because the value of acceleration is not the same at all points in time.

To look at it from mathematical perspective, consider a particle with position $\mathbf{s}$. The velocity is $\mathbf{v} = \dot{\mathbf{s}}$, and the acceleration is $\mathbf{a} = \dot {\mathbf{v}} = \ddot{\mathbf{s}}$. If the particle moves in a circle, then $\mathbf{s} = (\cos \omega t, \sin \omega t)$, where $\omega$ is the angular velocity (rate of change of angle per unit time). For simplicity, let $\omega=1$, so $\mathbf{s} = (\cos t, \sin t)$.

Taking this component-wise, we have $s_x = \cos t$ for the $x$ coordinate. This $\cos$ term may be seen from the unit circle. So for the $x$-component, the velocity is $v_x = \frac{d}{dt} s_x = -\sin t$, and the acceleration is $a_x = \frac{d}{dt} v_x = -\cos t$. Clearly, the acceleration in the $x$-component is not constant, but changes over time. Similarly for the $y$-component we have $a_y = \frac{d^2}{dt^2} \sin t = -\sin t$, which is also not constant.

Thus the acceleration for a particle moving in a circle at angular velocity $\omega = 1$ is $\mathbf{a} = (-\cos t, -\sin t)$, which we see directly is not constant over time, but changes continuously (it is variable).

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  • $\begingroup$ I tried to adjust my answer a bit after writing. In the second paragraph, it should be clear. As long as we agree on a definition for the term "variable" (I used it as a synonym for "changing") there shouldnt be any confusion. $\endgroup$ – Sam Gallagher Feb 1 at 14:06
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Acceleration is the rate of change of velocity and velocity is a vector quantity which means that it has both a magnitude and a direction.

So there is an acceleration if:

  • the magnitude of the velocity will change but not its direction
  • the direction of the velocity will change but not its magnitude
  • both the magnitude and the direction of the velocity will change

In the context of uniform circular motion that is taken to mean the the magnitude of the velocity (speed) of the object does not change whereas its direction does so as it moves along a circular trajectory, the object is accelerating and the magnitude of this acceleration is constant but the direction of the acceleration is changing at a rate equal to the angular speed of the rotating object.

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  • $\begingroup$ So you are saying that the acceleration is constant but in my book it is written that the acceleration is variable. However, I am not convinced that the acceleration is variable. I think that it is uniform(as direction of velocity is changing at a uniform rate) $\endgroup$ – Kushagra Shukla Feb 1 at 11:48
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    $\begingroup$ @KushagraShukla I said the magnitude of the acceleration is constant. The direction of the acceleration changes at a constant (uniform) rate equal to the angular speed of the object. $\endgroup$ – Farcher Feb 1 at 11:53
  • $\begingroup$ You've shown perfectly well that there is an acceleration in uniform circular motion, but you haven't addressed the OP's question as to why that acceleration is not constant. $\endgroup$ – Nuclear Wang Feb 1 at 16:48
  • $\begingroup$ @NuclearWang Thank you. I have added a little extra to my answer. $\endgroup$ – Farcher Feb 1 at 17:00
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I think you are confusing uniform circular motion with uniform acceleration. The book you are talking about is right in saying that the acceleration is variable. This is because although the magnitude is not changing, the direction is constantly changing. Furthermore, the rate at which the direction is changing is constant (hence, the use of the word "uniform").

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Uniform circular motion can be well-illustrated by swinging a tennis ball on a string in a circle over your head at a constant speed. The tension on the string is what accelerates the ball, so that acceleration vector always points from the ball to the center of the circle. The tension in the string remains constant, so the acceleration has constant magnitude, but as the ball and string moves, the acceleration vector changes direction. The acceleration vector is considered variable because its direction changes, even though its magnitude is constant.

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It's only variable if you're using a fixed orthogonal coordinate system fixed to the ground. (As other answers have explained). But, if you choose a moving coordinate system fixed on the moving object, the acceleration is fixed. That is, choose axes Tangential and Radial, and the acceleration will be constant, in the radial direction.

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Consider an object moving around a clock face. When it is at the 3'O Clock position it is moving down, and when it is at the 9'O Clock position it is moving up.

Now, if this movement is a uniform circular motion then the speed (magnitude of the velocity) is constant, but the velocity is constantly changing because, as you say, it is constantly changing direction.

Now consider the 3'O Clock position when it is moving downwards again. At this point it is being accelerated to the left. This results in a change of velocity so that it is moving less quickly downwards, and slightly to the left. Likewise when it is at the 9'O Clock position, it is being accelerated to the right, resulting in a decrease in how quickly it moves up, and an increase in how quickly it moves to the right (which it was not doing at all until this point, indeed just before it reached the 9'O Clock position it had a slight leftwards motion).

Hence just as the velocity is of constant magnitude (the speed) but constantly variable direction, so too the acceleration is also of constant magnitude but constantly variable direction (always at 90° to the motion). Because acceleration is a vector, just as velocity is, if you change the direction then you have changed it.

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