Tesla3
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Hi,
I can't seem to get how my book is deriving the acceleration in circular motion. The thing I don't get is what does the book means by saying that " the relation of the acceleration to the velocity is the same as the relation of the velocity to the displacement" I know that:

" v = u + a*t "
&
" s = u*t +(1/2)*a*(t^2) "

But still can't figure out in what sense the book is trying to imply the relation similarity to finally deduce the acceleration in circular motion. Can anyone please clarify the situation for me? I have attached my book's derivation method. Thanks in advance.
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K-Man_PhysCheM
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(Original post by Tesla3)
Hi,
I can't seem to get how my book is deriving the acceleration in circular motion. The thing I don't get is what does the book means by saying that " the relation of the acceleration to the velocity is the same as the relation of the velocity to the displacement" I know that:

" v = u + a*t "
&
" s = u*t +(1/2)*a*(t^2) "

But still can't figure out in what sense the book is trying to imply the relation similarity to finally deduce the acceleration in circular motion. Can anyone please clarify the situation for me? I have attached my book's derivation method. Thanks in advance.
Note you can't use the SUVAT equations here because they are only for constant linear acceleration, while in circular motion the direction is constantly changing. (In fact, really you should forget the SUVAT equations even exist: this is often one of the first things they tell you to do in first year undergrad physics. You should always do kinematics by integrating the equation of motion, though at A-level they write questions where it's much quicker to have just memorised SUVAT...).

To understand their argument, you just need to think geometrically. I think they have worded it as clearly as possible, so I don't really know what I can add, but I'll give it a go.

\mathbf{r} vector rotates at rate \omega and it's magnitude is r, so the magnitude of \mathbf{v} is \omega r. You should already know this.

Then, by the same token, since the \mathbf{v} vector rotates at a rate \omega as well, then the magnitude of the angular acceleration must be \omega times the magnitude of the velocity vector. Hence its magnitude is (\omega r) \omega = \omega^2 r, for exactly the same reason that the magnitude of \mathbf{v} is \omega r.


These geometric derivations are always tricky to get your head around. It becomes much easier to understand this when you do it "properly" (i.e. when you parametrise \mathbf{r} and differentiate each component and the basis vectors.)
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K-Man_PhysCheM
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I'll write out the parametrisation argument explicitly because you might find it easier to understand.

For motion around a circle of radius r at angular speed \omega, we can write

\mathbf{r}(t) = r \cos(\omega t) \mathbf{i} + r \sin(\omega t) \mathbf{j}.

As time increases, the vector \mathbf{r}(t) rotates anticlockwise in a circle of radius r about the origin, at an angular speed \omega.

The magnitude of this vector is found by Pythagoras' theorem as usual:

|\mathbf{r}(t)| = \sqrt{r^2 \cos^2(\omega t) + r^2 \sin^2(\omega t)} = r,

where we have used a well-known trig identity.

We obtain velocity and acceleration vector by differentiating each component once or twice respectively with respect to time.

\mathbf{v}(t) = \frac{\mathrm{d}\mathbf{r}(t)}{ \mathrm{d}t} = -r \omega \sin(\omega t) \mathbf{i} + r \omega\cos(\omega t) \mathbf{j}.

\mathbf{a}(t) = \frac{\mathrm{d}^2\mathbf{r}(t)}  {\mathrm{d}t^2} =  -r \omega^2 \cos(\omega t) \mathbf{i} - r \omega^2\sin(\omega t) \mathbf{j} = -\omega^2 \mathbf{r}(t).

You can easily check for yourself that the magnitudes are r\omega and r\omega^2 respectively!

(Original post by Tesla3)
Hi,
I can't seem to get how my book is deriving the acceleration in circular motion. The thing I don't get is what does the book means by saying that " the relation of the acceleration to the velocity is the same as the relation of the velocity to the displacement" I know that:

" v = u + a*t "
&
" s = u*t +(1/2)*a*(t^2) "

But still can't figure out in what sense the book is trying to imply the relation similarity to finally deduce the acceleration in circular motion. Can anyone please clarify the situation for me? I have attached my book's derivation method. Thanks in advance.
Last edited by K-Man_PhysCheM; 6 months ago
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Tesla3
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(Original post by K-Man_PhysCheM)
Note you can't use the SUVAT equations here because they are only for constant linear acceleration, while in circular motion the direction is constantly changing. (In fact, really you should forget the SUVAT equations even exist: this is often one of the first things they tell you to do in first year undergrad physics. You should always do kinematics by integrating the equation of motion, though at A-level they write questions where it's much quicker to have just memorised SUVAT...).

To understand their argument, you just need to think geometrically. I think they have worded it as clearly as possible, so I don't really know what I can add, but I'll give it a go.

\mathbf{r} vector rotates at rate \omega and it's magnitude is r, so the magnitude of \mathbf{v} is \omega r. You should already know this.

Then, by the same token, since the \mathbf{v} vector rotates at a rate \omega as well, then the magnitude of the angular acceleration must be \omega times the magnitude of the velocity vector. Hence its magnitude is (\omega r) \omega = \omega^2 r, for exactly the same reason that the magnitude of \mathbf{v} is \omega r.


These geometric derivations are always tricky to get your head around. It becomes much easier to understand this when you do it "properly" (i.e. when you parametrise \mathbf{r} and differentiate each component and the basis vectors.)
I get the v = r * omega in the first diagram.

For the second diagram,
we have R = v
and Omega = omega
V= V
Using the formula: " V = R * Omega "
we have " V = v * omega "
then putting " v = r*omega "
we have " V = r*(omega^2) "
But where does acceleration comes from? K-Man_PhysCheM
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K-Man_PhysCheM
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(Original post by Tesla3)
I get the v = r * omega in the first diagram.

For the second diagram,
we have R = v
and Omega = omega
V= V
Using the formula: " V = R * Omega "
we have " V = v * omega "
then putting " v = r*omega "
we have " V = r*(omega^2) "
But where does acceleration comes from? K-Man_PhysCheM
In the second diagram, "r" corresponds to v, and "v" corresponds to a.

So "a = v* omega"
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Tesla3
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(Original post by K-Man_PhysCheM)
In the second diagram, "r" corresponds to v, and "v" corresponds to a.

So "a = v* omega"
I see but can you tell me that what was it trying to say in the start of the page "The relation of the acceleration to the velocity is the same as the relation of the velocity to the displacement". What does this mean? I know its not required, just for the sake of information. Thanks btw
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K-Man_PhysCheM
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(Original post by Tesla3)
I see but can you tell me that what was it trying to say in the start of the page "The relation of the acceleration to the velocity is the same as the relation of the velocity to the displacement". What does this mean? I know its not required, just for the sake of information. Thanks btw
It means that the relation between velocity and displacement vectors is the same as the relation between acceleration and velocity vectors.

I.e. they have the same ratio of magnitudes and are always at the same angle to each other (at right angles).

This geometric proof feels almost a bit hand-wavey, but it is logically sound since the vectors are always at right-angles to each other and so rotate at the same rate and are related to each other in the same way (by differentiating), hence are just scalings of each other. Do take a look at the 2nd post I made where I detailed the more "standard" proof, which generalises to circular motion at non-constant speed.
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Tesla3
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(Original post by K-Man_PhysCheM)
It means that the relation between velocity and displacement vectors is the same as the relation between acceleration and velocity vectors.

I.e. they have the same ratio of magnitudes and are always at the same angle to each other (at right angles).

This geometric proof feels almost a bit hand-wavey, but it is logically sound since the vectors are always at right-angles to each other and so rotate at the same rate and are related to each other in the same way (by differentiating), hence are just scalings of each other. Do take a look at the 2nd post I made where I detailed the more "standard" proof, which generalises to circular motion at non-constant speed.
Sorry for asking another amateur question, but is the angular acceleration " a " same for both the diagram one and two?
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(Original post by Tesla3)
Sorry for asking another amateur question, but is the angular acceleration " a " same for both the diagram one and two?
These aren't amateur questions! It's a tricky topic to get your head around the first time you meet it!

"a" isn't angular acceleration, it is the linear acceleration vector. Fig 6.4 and Fig 6.5 are diagrams of the same situation. You could combine them into one diagram, where you put \mathbf{a} on the end of vector \mathbf{v} in Fig 6.4

The angular acceleration \alpha is zero for constant-speed circular motion. The angular acceleration would only be non-zero if the angular velocity, \omega, changed over time, since \alpha is the derivative of \omega with respect to time.
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Tesla3
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(Original post by K-Man_PhysCheM)
These aren't amateur questions! It's a tricky topic to get your head around the first time you meet it!

"a" isn't angular acceleration, it is the linear acceleration vector. Fig 6.4 and Fig 6.5 are diagrams of the same situation.

The angular acceleration \alpha is zero for constant-speed circular motion. The angular acceleration would only be non-zero if the angular velocity, \omega, changed over time.
oh sorry, I meant to say that whether the radial acceleration for the first diagram is equivalent to the the transversal acceleration of the second diagram?
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(Original post by Tesla3)
oh sorry, I meant to say that whether the radial acceleration for the first diagram is equivalent to the the transversal acceleration of the second diagram?
Yes, that's the whole point of the 2nd diagram!
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Tesla3
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(Original post by K-Man_PhysCheM)
Yes, that's the whole point of the 2nd diagram!
I see, but how are we so sure about the fact that there radial acceleration of the first diagram will be equivalent to the linear velocity in the second diagram.
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(Original post by Tesla3)
I see, but how are we so sure about the fact that there radial acceleration of the first diagram will be equivalent to the linear velocity in the second diagram.
Because all three vectors, \mathbf{r}, \mathbf{v} and \mathbf{a} are always at right-angles to each other when \omega is constant, and so all three rotate at the same rate, \omega.

They are fixed relative to each other by the constraint that they are always at right-angles to each other.
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(Original post by K-Man_PhysCheM)
Because all three vectors, \mathbf{r}, \mathbf{v} and \mathbf{a} are always at right-angles to each other when \omega is constant, and so all three rotate at the same rate, \omega.

They are fixed relative to each other by the constraint that they are always at right-angles to each other.
I get that the direction of linear velocity in the second diagram will be equivalent to the radial acceleration of the first diagram, but what about the magnitude, how are we sure that the magnitude of the radial acceleration in the first diagram is equivalent to the magnitude of the linear velocity in the second diagram.
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K-Man_PhysCheM
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(Original post by Tesla3)
I get that the direction of linear velocity in the second diagram will be equivalent to the radial acceleration of the first diagram, but what about the magnitude, how are we sure that the magnitude of the radial acceleration in the first diagram is equivalent to the magnitude of the linear velocity in the second diagram.
We get the magnitude from the fact that they are all rotating at the same angular speed, \omega
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(Original post by K-Man_PhysCheM)
We get the magnitude from the fact that they are all rotating at the same angular speed, \omega
I think, ummmm, still getting a bit confused, I know they are rotating at the same angular speed omega, that's not a problem, but how do we deduce from omega that the magnitude is same. Sorry for bothering you again and again.......
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(Original post by Tesla3)
I think, ummmm, still getting a bit confused, I know they are rotating at the same angular speed omega, that's not a problem, but how do we deduce from omega that the magnitude is same. Sorry for bothering you again and again.......
In the same way that you deduce that the magnitude of v is omega times the magnitude of r.
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Tesla3
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(Original post by K-Man_PhysCheM)
In the same way that you deduce that the magnitude of v is omega times the magnitude of r.
wait let me think........
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Tesla3
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Can we use this fact to help us out in some way:

|a|/|v|=|s|/|v|

|a|/(r*omega) = (r*theta)/(r*omega)

|a| = r * theta

but what will we do further?
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K-Man_PhysCheM
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(Original post by Tesla3)
Can we use this fact to help us out in some way:

|a|/|v|=|s|/|v|

|a|/(r*omega) = (r*theta)/(r*omega)

|a| = r * theta

but what will we do further?
You have the ratios the wrong way round.

|a|/|v|=|v|/|s|

since the relabelling when we go Fig 6.4 --> Fig 6.5 is
r --> v, and
v --> a

so |r|/|v| = |v|/|a|

Then |a|/|v| = |v|/|r| = omega
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