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I recently came across this diagram in my physics textbook

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I got a bit confused by this diagram. I thought that the reason why the object accelerates towards the center is because of a net force, not because of velocities being added together.

So my question is: what forces do we use to determine that the centripetal force goes towards the center? I thought that we would use forces, and not velocities. Could you point out where I am confused?

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  • $\begingroup$ The text and diagrams above attempt to explain that if a particle follows a circular path at a constant rate, then the particle's acceleration must have constant magnitude, and it must point toward the center of the circle at all times. The text and diagrams say nothing about "the reason why the object accelerates..." Common reasons for things actually moving in a circle include, that the object in question is part of or attached to some larger, rigid, rotating body; and the object in question is orbiting a planet or a star. $\endgroup$ – Solomon Slow Apr 4 at 15:15
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$\bullet$ Note that the direction of the change in the velocity $\Delta\textbf{v}$ is towards the center. Therefore, the acceleration must be towards the center and hence also the force. Your diagram conveys the point that if a circular motion is uniform i.e., the magnitude of velocity remains fixed while its direction changes, the motion must be due to a force solely directed towards the center.

$\bullet$ The centripetal force may be provided by different agent in different situations. For the motion of planets around the sun, the centripetal force is provided by the gravitational attraction of the sun, for a stone tied to a string rotated in a circle the centripetal force is provided by the tension in the string and so on.

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  • $\begingroup$ How does it show that the magnitude of the velocity remains fixed? Wouldn't subtracting final velocity and initial velocity give some value which shows the velocity is changing? $\endgroup$ – Christopher Apr 4 at 9:01
  • $\begingroup$ Your diagram doesn't clearly show it except drawing arrows of equal length. If and only if the magnitude of velocity is fixed for a circular motion, the net force be directed towards the centre. $\endgroup$ – SRS Apr 4 at 9:05
  • $\begingroup$ So is there a change in velocity? I thought that the velocity would be constant, but wouldn't doing V final minus V initial give you a value, which would indicate the velocity is changing? I'm just confused as to why they do Vf - Vi and give it a magnitude when I thought it should be zero $\endgroup$ – Christopher Apr 4 at 9:06
  • $\begingroup$ Yes. Velocity is a vector quantity. Even if its magnitude stays constant but direction changes you'll have a nonzero acceleration. This is exactly what happens in a uniform circular motion where $\textbf{v}=R\omega\hat{r}$ where $R$ is the radius of the circle and $\hat{r}$ is a unit vector that points radially outward. Though $R\omega$ is a constant the unit vector $\hat{r}$ changes from one instant to another. $\endgroup$ – SRS Apr 4 at 9:11
  • $\begingroup$ Right, so the change in velocity when you consider direction is not zero, but the magnitude of the velocities would subtract to zero? $\endgroup$ – Christopher Apr 4 at 9:14
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The diagrams show that if a body is moving in a circular path at constant speed, then its acceleration (rate of change of velocity) must be towards the centre of the circle. To have this acceleration, Newton's second law tells us that the body must be experiencing a resultant force always directed towards the centre of the circle. That's the logic.

So you could indeed say that "the reason why the object accelerates towards the center is because of a net force". But the vector stuff shows why the resultant force must be directed towards the circle centre if the body is to move in a circle. It also tells you the magnitude ($mv^2/r$) of resultant force needed for a circular path of radius $r$ at a particular speed $v$, since it shows that the magnitude of the acceleration must be $v^2/r$.

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  • $\begingroup$ Thanks for the answer. This question may sound stupid, but how do we determine the net force? Does doing that vector subtraction stuff find the resultant force, or is it more complicated than that? Because I thought that when determining net forces, we use forces and not velocity. $\endgroup$ – Christopher Apr 4 at 8:43
  • $\begingroup$ The relation is simple, $\sum F=ma $ . Find the net acceleration of the object from its change in velocity, and find the net force. $\endgroup$ – Tojrah Apr 4 at 8:47
  • $\begingroup$ It depends what information you have. If you know the individual forces acting on a body, their vector sum gives the resultant force. But if you know that the body is moving in a circle of radius $r$ at constant speed $v$ then you can use vector subtraction of velocities and N's Law 2 to deduce that the resultant force has to be $mv^2/r$ always directed to the circle centre. $\endgroup$ – Philip Wood Apr 4 at 8:49
  • $\begingroup$ Someting else confused me as well. Doesn't the object move at constant velocity when in circular motion? Wouldn't subtracting V final and V initial give some value? $\endgroup$ – Christopher Apr 4 at 9:05
  • $\begingroup$ Velocity is a vector. Even if a body is moving in a circle at constant speed, its direction of motion keeps changing, so its velocity keeps changing. The diagrams you have displayed show that subtracting two velocities with the same magnitude, but with different directions, give a vector of non-zero magnitude. I recommend that you read up about vectors, and how to add and subtract them. You'll be using these techniques again and again in your study of Physics. $\endgroup$ – Philip Wood Apr 4 at 9:15
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I have annotated your diagram by labelling angles $\alpha$ and $\beta$ in the vector subtraction triangle.

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As the angle between the initial velocity vector and the final velocity vector, $\alpha$, becomes smaller and smaller the angle between the initial velocity vector and the change in velocity vector becomes closer and closer to a right angle.
In the limit of $\alpha$ tending to zero because the initial velocity vector is a tangent to the circle, the change in velocity must be towards the centre of the circle.
This means that the acceleration and hence the force causing this acceleration must point towards the centre of the circle.

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Force is proportional to acceleration (for constant mass). The diagram is calculating the change in velocity $\Delta \vec v$ for a small time interval. If you know the corresponding time interval $\Delta t$ then $\frac{\Delta \vec v}{\Delta t}$ gives you the average acceleration over this time interval. The final step, which is glossed over in the explanation, is that the limit of $\frac{\Delta \vec v}{\Delta t}$ as $\Delta t \rightarrow 0$ is an instantaneous acceleration vector with constant length that always points towards the centre of the circle.

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