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My maths teacher explained this to be me by way of analogy: a car driving around a perfectly circular track would be constantly changing its velocity (while the magnitude of the velocity is not changing, the direction is). Because acceleration is the rate of change of velocity, and the object is changing direction, it is said to be accelerating. This strikes me as an odd definition of acceleration, as surely it still equals $\mathrm{0 ms^{-2}}$, even if the object is changing direction. Nevermind, I thought, it's just a definition.

What is strange is that this seeming technicality actually tells us information. Because the object is accelerating, there must have been a resultant force acting on it (since $F=ma$). This is completely baffling to me — yes, the object must have been 'accelerating', but if the magnitude of that acceleration = $0\mathrm{ms^{-2}}$, it seems certain that $F$ = 0 as well. What am I missing?

Joe
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  • Because velocity and acceleration are vectors made of magnitude and direction. – Mauro ALLEGRANZA Sep 17 '19 at 16:16
  • Acceleration is change of velocity, and the change may affect either magnitude or direction (or both) – Mauro ALLEGRANZA Sep 17 '19 at 16:17
  • In the uniform circular motion we have change of direction only and in order to do it we need a force that produce continuously a change of direction (an acceleration). – Mauro ALLEGRANZA Sep 17 '19 at 16:18
  • @MauroALLEGRANZA I understand that the object must have been accelerating, but why does that mean a resultant force must have been acting on that object if the acceleration does not involve a change in magnitude? – Joe Sep 17 '19 at 16:18
  • Because we have the Law of Inertia : no external force, no change of velocity (vector). Thus, by contraposition, if yes change of velocity, then yes force. By Newton's 2nd Law, if force, then acceleration. It is the foundation of Newtonian mdynamics. – Mauro ALLEGRANZA Sep 17 '19 at 16:23
  • @MauroALLEGRANZA If an object's acceleration = $0$ though (as it is only accelerating in the sense that it is changing direction), then plugging this is into the equation $F=ma$ gives a resultant force of $0$. So how is the resultant force merely a change in direction too? – Joe Sep 17 '19 at 16:31
  • See Uniform circular motion : "Since the object's velocity vector is constantly changing direction, the moving object is undergoing acceleration by a centripetal force in the direction of the center of rotation. Without this acceleration, the object would move in a straight line, according to Newton's laws of motion.". The acceleration is not $0$. – Mauro ALLEGRANZA Sep 17 '19 at 16:34
  • For a body moving in a circle of radius $r$ with angular velocity $\omega$ we have : $a=\frac {v^{2}}{r}=\omega ^{2}r$. – Mauro ALLEGRANZA Sep 17 '19 at 16:37
  • "This strikes me as an odd definition of acceleration, as surely it still equals $0ms^{−2}$, even if the object is changing direction. " Why is that "surely". The direction is changing so surely that is not $0ms^{-2}$. It would need a directional variable as well. – fleablood Sep 17 '19 at 16:56
  • @MauroALLEGRANZA I just put up an answer (now deleted) referring to the law of inertia - essentially what you said. Then I saw your comment. Always good to make sure fundamental principles are understood. – Mark Bennet Sep 17 '19 at 17:19
  • You’re confusing velocity, which is a vector, with speed, which is a scalar equal to the magnitude of the velocity. – amd Sep 17 '19 at 17:27
  • If we are considering velocity in two rather than one dimension is is a two variable pair value. $(r, \theta)$ where $r$ is a non-negative magnitude and $\theta$ is an angular direction. Acceleration will be a $(\Delta r, \Delta \theta)$ pair. If the car is driving in a circle then $(\Delta r, \Delta \theta) = (0, \alpha)$ where the magnitude of the velocity never changes (and is thus $0$) but the direction of the velocity changes at a constant $\alpha\ne 0$ rate. $(0,\alpha) \ne 0$. – fleablood Sep 17 '19 at 17:28

3 Answers3

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Acceleration means the rate of change of velocity. Here velocity is a vector which can have the same magnitude in different directions. You have considered only the magnitude, not the direction.

$$\overrightarrow{a}=\frac{\overrightarrow{v}-\overrightarrow{u}}{t}$$

Here $a$ is average acceleration. Consider an object moving in x-y plane along a circular path of radius $1$ centred at the origin with a constant speed $2 m/s$. Velocity of it at $(1,0)$ is tangential to the path and is $2j$. At point $(0,1)$ velocity is tangential to the path and is $-2i$. Time taken by it to travel between these two points is $(distance/speed)$, i.e $π/4$. so average acceleration is:

$$\overrightarrow{a}=\frac{-2i+2j}{\pi/4}\neq0$$

Jibin N Sunil
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  • You can learn about writing math expressions here: https://math.meta.stackexchange.com/q/5020/504401 –  Oct 24 '19 at 12:15
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If it is changing direction, then its motion is changing too. This is intuitively what we understand by acceleration. For example, if you were in a vehicle which is changing its direction only, you would feel the changes in the motion of the vehicle.

It's the change in motion (velocity) that we call acceleration.

Allawonder
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ACCELEARTION IS A VECTOR QUANTITY!!!

What your Maths Teacher may have meant is that the object changing direction has different acceleration , in the sense that the magnitude of the acceleration is constant (provided no external force) ; it is only changing it's resultant direction .

In Vector form , A Vector A can be written in terms of it rectangular components :

Hence , $A = Acos\theta + Asin\theta $ , Where $\theta $ Is The angle of direction

Since the object is changing direction , at other instant , let the angle made be $\phi$

Then $A' = Acos\phi + A\sin\phi$

Clearly A is not equal to A' Although magnitude of $A = \sqrt{(Acos\theta)^2 +(Asin\theta)^2 } = A$

And magnitude of $A' = \sqrt{(Acos\phi)^2 +(Asin\phi)^2 } = A$

Both acceleration vector have the same magnitude but acts along different direction, so acceleration is changing.

The Demonix _ Hermit
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