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Recently I asked this question that how perpendicular acceleration can change the direction of velocity without changing its magnitude, one of the explanation was for a small interval of time that perpendicular acceleration will change the velocity of the object in the direction of acceleration that is given by this equation $$V_i + a \ dt = V_f$$ (where $V_i$ and $V_f$ are the initial and final velocity of the particle and $a$ is the perpendicular acceleration) as the gain in velocity(magnitude) is very small and the force acts on the particle in that particular direction for a very small time dt and also it is a differential quantity it can be ignored my doubt is:-

1)The change in direction is also very small why cant we ignore it means how can adt change the direction of velocity without changing magnitude especially if acceleration is perpendicular to velocity.

2)That increase in magnitude is a small quantity but it is getting added at every instance so its integration is non zero. So why are we ignoring it.

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$$|\mathbf v|^2=\mathbf v\cdot \mathbf v$$ $$\frac{\text d}{\text dt}\left(|\mathbf v|^2\right)=2\mathbf v\cdot\frac{\text d\mathbf v}{\text dt}$$

As you can see, if the acceleration $\mathbf a=\text d\mathbf v/\text dt=0$, then the magnitude of the velocity vector $\mathbf v$ will not change.

That increase in magnitude is a small quantity but it is getting added at every instance so its integration is non zero. So why are we ignoring it.

You're constraining the acceleration to always be perpendicular to the velocity. So the "small quantity" is in a different direction at each instant, as the velocity direction changes at each instant. You are right that if you only had the acceleration perpendicular to the velocity at only one instant and then kept the acceleration fixed there, then the velocity would also change magnitude at future times. (For example, think of the velocity of a projectile at the top of its trajectory and then as it starts to fall back down).

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The vector addition $\vec v_{\rm f} = \vec v_{\rm i} + \vec a \,\Delta t$ with the acceleration $\vec a$ perpendicular to the initial velocity $\vec v_{\rm i}$ is shown below.

enter image description here

From the diagram $v_{\rm f} = v_{\rm i}\,\cos \Delta \theta + a \,\Delta t \,\sin \Delta \theta$.
Consider what happens as $\Delta t$ and hence $\Delta \theta$ get smaller and smaller, $\cos \Delta \theta \to 1$ and $\sin \Delta \theta \to \Delta \theta$.

Thus $v_{\rm f} \to v_{\rm i}\cdot 1+a\,\Delta t \,\Delta \theta$

As $\Delta t$ (and $\Delta \theta$) get smaller and smaller the term $a\,\Delta t \,\Delta \theta$ containing the product of two quantities getting smaller and smaller can be neglected compared with the term $a\,\Delta t\cdot \cos \Delta \theta\to a\,\Delta t$ which contains only one quantity getting smaller and smaller.

So in summary, an acceleration at right angles to direction of a velocity can result in the direction of the velocity changing but without a change in the magnitude of the velocity.

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  • $\begingroup$ Thank you very much this was my doubt from two years I always skip this part and accept this as a fact and move on but this time I reviewed this part once again and this time my doubt was cleared. $\endgroup$ Commented Oct 12, 2022 at 13:52
  • $\begingroup$ @farcher,a small correction,I think you intended to write vfinal=vinitial(cos......) $\endgroup$ Commented Mar 24, 2023 at 17:31
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    $\begingroup$ @DheerajGujrathi Well spotted and I have made the correction. $\endgroup$
    – Farcher
    Commented Mar 25, 2023 at 0:01
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The equation you are thinking of isn't actually $$V_i + a \ dt = V_f$$ but rather $$\vec V_i + \vec a \ \Delta t = \vec V_f$$ Now, the second equation is only valid if $\vec a$ is constant over the time interval $\Delta t$, which is not the case, but we will take care of that later by taking the limit as $\Delta t \to 0$.

Since you are specifying that $\vec a$ is perpendicular to $\vec V_i$ we can specify (without further loss of generality) that $\vec V_i = (V_{xi},0)$ and $\vec a =(0,a_y)$ so that $$\vec V_f=(V_{xi}, a_y \Delta t)$$ We are interested in the change in the magnitude and the change in the direction. So the change in direction is given by $$\Delta \theta=\tan^{-1}\left(\frac{a_y \Delta t}{V_{xi}}\right)-\tan^{-1}\left(\frac{0}{V_{xi}}\right)=\frac{a_y}{V_{xi}}\Delta t-\frac{a_y^3}{3 V_{xi}^3}\Delta t^3 + ...$$ and the change in magnitude is given by $$\Delta |\vec V|=\sqrt{a_y^2 \Delta t^2+V_{xi}^2}-\sqrt{V_{xi}^2}=\frac{a_y^2}{2 V_{xi}}\Delta t^2-\frac{a_y^4}{8 V_{xi}^3}\Delta t^4+...$$

Now, as you note in your question $|\vec V|>0$, but that is only because we falsely treated $\vec a$ as though it were constant over $\Delta t$ when in fact $\vec a$ changes to remain perpendicular. To fix that, we need to take the limit as $\Delta t \to 0$ as follows: $$\frac{d\theta}{dt}=\lim_{\Delta t \to 0}\frac{\Delta \theta}{\Delta t}=\lim_{\Delta t \to 0}\left(\frac{a_y}{V_{xi}}-\frac{a_y^3}{3 V_{xi}^3}\Delta t^2 + ...\right)=\frac{a_y}{V_{xi}}$$$$ \frac{d|\vec V|}{dt}=\lim_{\Delta t \to 0}\frac{\Delta |\vec V|}{\Delta t} = \lim_{\Delta t \to 0} \left( \frac{a_y^2}{2 V_{xi}}\Delta t-\frac{a_y^4}{8 V_{xi}^3}\Delta t^3+... \right)=0 $$

1)The change in direction is also very small why cant we ignore it means how can adt change the direction of velocity without changing magnitude especially if acceleration is perpendicular to velocity.

The issue is the order of smallness. $\Delta \theta$ has a $\Delta t$ term, so when we take the limit that term does not disappear.

2)That increase in magnitude is a small quantity but it is getting added at every instance so its integration is non zero. So why are we ignoring it.

Again, it is the order of smallness. $\Delta |\vec V|$ has the largest term is a $\Delta t^2$ term, so when we take the limit it goes to 0. In other words, as you say, the change in direction is very small (1st order), but the change in magnitude is very very small (2nd order). The very small quantities do not go away, but the very very small ones do.

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    $\begingroup$ Thank you for your beautiful explanation, I am also thinking the same way but I don't know enough mathematics to go into the detail I was already convinced by @Farcher but your answer made it more clear $\endgroup$ Commented Oct 12, 2022 at 14:36

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