Acceleration between two frames

[Monsterboy]

Homework Statement

The accelerations of a particle as seen from two frames S1 and S2 have equal magnitude $4 m/s^2$.

(a) The frames must be at rest with respect to each other.
(b) The frames may be moving with respect to each other but neither should be accelerated with
respect to the other.
(c) The acceleration S2 with respect to S1 may either be zero or $8 m/s^2$.
(d) The acceleration of S2 with respect to S1 may be anything between zero and $8 m/s^2$.

Relevant Equations

$v_{P,S^1} = v_{P,S^2} + v_{S^1,S^2}$ Velocity of the particle with respect to the frames and velocity of one frame with respect to another.
$a_{P,S^1} = a_{P,S^2} + a_{S^1,S^2}$ Acceleration of the particle with respect to the frames and acceleration of one frame with respect to another.

The correct option is given as (d)

I think I am able to visualize the problem but not able to put it in the equations shared above.

If the the two frames are moving away from the particle at $4 m/s^2$ in opposite directions we get the acceleration between the frames as $8 m/s^2$.
Substituting in the above equation for acceleration, we get 4 = -4 + 8, is this right ?

If the two frames are moving at constant velocity in opposite directions we get 4 = 4 + 0 right ? from the above equation for acceleration.

When the two frames are moving away from the particle at $4 m/s^2$ at any angle other than $180^0$ or zero degrees, how do I represent that in an equation ?

 

[PeroK]

When the two frames are moving away from the particle at $4 m/s^2$ at any angle other than $180^0$ or zero degrees, how do I represent that in an equation ?

Acceleration is a vector.

 

[Monsterboy]

Acceleration is a vector.

Yea, is there some kind of a mathematical way to represent that ? I mean the two frames moving away from the particle at 4 $m/s^2$ at an angle $\theta$ with respect to each other ?

 

[PeroK]

Yea, is there some kind of a mathematical way to represent that ? I mean the two frames moving away from the particle at 4 $m/s^2$ at an angle $\theta$ with respect to each other ?

Use vectors.

 

[Monsterboy]

Use vectors.

You mean like this ? $\vec{a}_{P,S^1} = \vec{a}_{P,S^2} + \vec{a}_{S^1,S^2}$ is that it ?

 

[PeroK]

You mean like this ? $\vec{a}_{P,S^1} = \vec{a}_{P,S^2} + \vec{a}_{S^1,S^2}$ is that it ?

Not quite - I assume that $\vec{a}_{S^1,S^2}$ means that acceleration of $S^1$ in frame $S^2$. If so, shouldn't that be the other way round?

 

[Monsterboy]

Not quite - I assume that $\vec{a}_{S^1,S^2}$ means that acceleration of $S^1$ in frame $S^2$. If so, shouldn't that be the other way round?

By $\vec{a}_{S^1,S^2}$ I meant acceleration of frame $S^1$ with respect to frame $S^2$.

 

[PeroK]

By $\vec{a}_{S^1,S^2}$ I meant acceleration of frame $S^1$ with respect to frame $S^2$.

Yes, that what I thought. That's the wrong way round. Draw a diagram with the point $P$ origin of $S^2$ in the frame of $S^1$.

 

[Monsterboy]

Yes, that what I thought. That's the wrong way round. Draw a diagram with the point $P$ origin of $S^2$ in the frame of $S^1$.

Oh, you mean $\vec{a}_{P,S^1} = \vec{a}_{P,S^2} + \vec{a}_{S^2,S^1}$

 

[PeroK]

Oh, you mean $\vec{a}_{P,S^1} = \vec{a}_{P,S^2} + \vec{a}_{S^2,S^1}$

Yes. Doesn't that make more sense? It's the same for relative acceleration as it is for relative velocity and relative position. You start with the formula for relative position and differentiate wrt time.

 

[Monsterboy]

Yes. Doesn't that make more sense? It's the same for relative acceleration as it is for relative velocity and relative position. You start with the formula for relative position and differentiate wrt time.

Yea, so if I put $\vec{a}_{P,S^1} =4 m/s^2 = \vec{a}_{P,S^2}$ and let's say $\vec{a}_{S^2,S^1} = 4 m/s^2$. Can I find $\theta$ i.e angle between the two frames ?

How should I proceed ?

 

[PeroK]

Yea, so if I put $\vec{a}_{P,S^1} =4 m/s^2 = \vec{a}_{P,S^2}$ and let's say $\vec{a}_{S^2,S^1} = 4 m/s^2$. Can I find $\theta$ i.e angle between the two frames ?

How should I proceed ?

What about looking at the acceleration of the two frames relative to the particle? What do you know about the two acceleration vectors $a_{s^1, P}$ and $a_{s^2, P}$?

 

[kuruman]

I think that this problem is ill-posed. I have a counterexample that matches the parameters of the problem but results in all four choices being false. I will post it after this thread has run its course but anyone who cannot wait (other than @Monsterboy) may ask me via PM.

 

[Monsterboy]

What about looking at the acceleration of the two frames relative to the particle? What do you know about the two acceleration vectors $a_{s^1, P}$ and $a_{s^2, P}$?

The magnitudes of $a_{S^1, P}$ and $a_{S^2, P}$ both are $4 m/s^2$ right ?
because $a_{P, S^1}$ and $a_{P, S^2}$ are both $4 m/s^2$, I am assuming the particle to be at rest and the two frames to be accelerating at $4 m/s^2$ with respect to the particle, at an angle $\theta$ between each other.

So, if $a_{S^2, S^1}$ is $4 m/s^2$. We get a kind of triangle with one vertex being the particle $P$ and two other vertices being the two frames $S^1$ and $S^2$. and the lengths $PS^1$ and $PS^2$ are extending at the rate of $4 m/s^2$. The length $S^1 S^2$ should also extend at the rate of $4 m/s^2$ right ?

Am I visualizing this correctly ?

 

[PeroK]

Am I visualizing this correctly ?

I don't think so. If the magnitude of the acceleration is $4m/s$, what does that tell you about the acceleration vector? Perhaps think about position as an analogy. If the position (vector) has a magnitude of $4m$, say, what does that tell you about the position relative to the origin?

 

[Monsterboy]

So, my expanding triangle analogy is wrong ?

Perhaps think about position as an analogy. If the position (vector) has a magnitude of $4m$, say, what does that tell you about the position relative to the origin?

The position vector will have a magnitude of $4m$ pointed away from the origin ?

If the magnitude of the acceleration is 4m/s, what does that tell you about the acceleration vector?

You mean $\vec{a}_{S1, P}$ ? It is pointed away from the particle P ?

 

[vela]

So, if $a_{S^2, S^1}$ is $4 m/s^2$. We get a kind of triangle with one vertex being the particle $P$ and two other vertices being the two frames $S^1$ and $S^2$. and the lengths $PS^1$ and $PS^2$ are extending at the rate of $4 m/s^2$. The length $S^1 S^2$ should also extend at the rate of $4 m/s^2$ right ?

Am I visualizing this correctly ?

You can visualize it that way, but it's not very helpful in solving the problem. And the conclusion you drew is wrong too. It seems like you're thinking of similar triangles, i.e., if the length of two sides of a triangle double, then the third side doubles too, which doesn't really apply here.

You have the vector equation relating the three accelerations. Draw the corresponding figure that represents that equation.

 

[PeroK]

The position vector will have a magnitude of $4m$ pointed away from the origin ?

You mean $\vec{a}_{S1, P}$ ? It is pointed away from the particle P ?

It's lies on a circle (2D) or sphere (3D) of radius $4m$. Never forget that the circle and sphere define the points of common distance from the origin hence common magnitude of position vector.

 

[haruspex]

I feel the wording of (d) is not quite what is intended.
As written, its truth depends on whether, for any given value in (0,8)m/s2, there is an arrangement which satisfies the criteria. It does not depend on the feasibility of any other values.
Better would have been "[0,8] and only [0,8]".
Certainly the solver should show that all values in (0,8) are achievable.

 

[Monsterboy]

You can visualize it that way, but it's not very helpful in solving the problem. And the conclusion you drew is wrong too. It seems like you're thinking of similar triangles, i.e., if the length of two sides of a triangle double, then the third side doubles too, which doesn't really apply here.

You have the vector equation relating the three accelerations. Draw the corresponding figure that represents that equation.

That is the figure that I could draw. You have two frames $S1$ and $S2$ and a particle $P$. If you randomly place 3 points in 3D space or a 2D plane, they always form a triangle together unless they are in a straight line, right ?

$\vec{a}_{S1, P} = \vec{a}_{S2, P} = 4m/s^2$ so I assumed that the two frames are accelerating away from the particle.

 

[vela]

That is the figure that I could draw. You have two frames $S1$ and $S2$ and a particle $P$. If you randomly place 3 points in 3D space or a 2D plane, they always form a triangle together unless they are in a straight line, right?

Consider the vector equation $\vec A + \vec B = \vec C$. How do you add two vectors graphically? Review the chapter on vectors if you don't remember.

Once you have the picture, erase the label $\vec C$ and write $\vec a_{P,S1}$, erase the label $\vec A$ and write $\vec a_{P,S2}$, and erase the label $\vec B$ and write $\vec a_{S2,S1}$.

$\vec{a}_{S1, P} = \vec{a}_{S2, P} = 4m/s^2$ so I assumed that the two frames are accelerating away from the particle.

Just a remark on notation. It doesn't make sense to write something like $\vec a = 4~\rm m/s^2$. The lefthand side is a vector quantity while the righthand side is a scalar. They're different types of mathematical objects. The correct notation is $\| \vec a \| = 4~\rm m/s^2$, or it's common to write $a = 4~\rm m/s^2$ (notice the lack of an arrow) where $a$ is understood to mean "the magnitude of $\vec a$".

 

[Monsterboy]

Consider the vector equation $\vec A + \vec B = \vec C$. How do you add two vectors graphically? Review the chapter on vectors if you don't remember.

Once you have the picture, erase the label $\vec C$ and write $\vec a_{P,S1}$, erase the label $\vec A$ and write $\vec a_{P,S2}$, and erase the label $\vec B$ and write $\vec a_{S2,S1}$.

So, you can add the vectors by parallelogram law or triangle law, the two vectors $\vec A , \vec B$ are like two adjacent sides of a parallelogram and the $\vec C$. is like the diagonal.

$\vec A + \vec B = \vec C$ will be $\vec a_{P,S2} + \vec a_{S2,S1} = \vec a_{P,S1}$

we get $tan \alpha = \frac{a_{S2, S1}}{a_{P, S1} + a_{P, S2}cos \theta}$

where $\theta$ is the angle between $\vec a_{P,S2}$ and $\vec a_{S2,S1}$ and $\alpha$ is the angle between $\vec a_{P,S1}$ and $\vec a_{S2,S1}$.

So, if the magnitude of all the above vectors is $4 m/s^2$ can I find the angle between the trajectories of the frames $S1$ and $S2$ (if this makes sense) ?

I was trying to visualize the above in 3D space and I thought it looked like a triangle.

Just a remark on notation. It doesn't make sense to write something like a→=4 m/s2. The lefthand side is a vector quantity while the righthand side is a scalar. They're different types of mathematical objects. The correct notation is ‖a→‖=4 m/s2, or it's common to write a=4 m/s2 (notice the lack of an arrow) where a is understood to mean "the magnitude of a→".

Yea, sorry I was using the notations carelessly.

 

[PeroK]

So, if the magnitude of all the above vectors is $4 m/s^2$ can I find the angle between the trajectories of the frames $S1$ and $S2$ (if this makes sense) ?

I was trying to visualize the above in 3D space and I thought it looked like a triangle.

Vector addition generally forms a triangle - unless the vectors are in the same direction, in which case it degenerates to a straight line.

If we take any two vectors of unit length, say, and add them together (in either 2D or 3D), then we can form a vector of any length from $0$ to $2$ units. You should be able to see that visually. If you want an algebraic confirmation you can use the law of cosines.

You also have the triangle inequaility. If $\vec A + \vec B = \vec C$, then $$|\vec C| \le |\vec A| + |\vec B|$$with equality only when $\vec A$ and $\vec B$ are in the same direction.

Again, using the triangle inequality, you should be able to see that $|\vec C|$ can be anything from $|\vec A| - |\vec B|$ to $|\vec A| + |\vec B|$, assuming that $\vec A$ is the larger of the two vectors.

 

[vela]

So, you can add the vectors by parallelogram law or triangle law, the two vectors $\vec A , \vec B$ are like two adjacent sides of a parallelogram and the $\vec C$. is like the diagonal.

$\vec A + \vec B = \vec C$ will be $\vec a_{P,S2} + \vec a_{S2,S1} = \vec a_{P,S1}$.

Good. You're interested in $\vec a_{S2,S1}$ which is equal to $\vec a_{P,S1} - \vec a_{P,S2} = \vec a_{P,S1} + (-\vec a_{P,S2})$. In your diagram, you simply need to reverse the direction of $\vec a_{P,S2}$. It's probably easier to see in a triangle diagram than in the parallelogram diagram.

Now consider how you can orient $\vec a_{P,S1}$ and $-\vec a_{P,S2}$ to maximize and minimize the magnitude of $\vec a_{S2,S1}$. Just fix the direction of $\vec a_{P,S1}$ in the x-direction and vary the direction of $-\vec a_{P,S2}$.

 

[Monsterboy]

Good. You're interested in $\vec a_{S2,S1}$ which is equal to $\vec a_{P,S1} - \vec a_{P,S2} = \vec a_{P,S1} + (-\vec a_{P,S2})$. In your diagram, you simply need to reverse the direction of $\vec a_{P,S2}$. It's probably easier to see in a triangle diagram than in the parallelogram diagram.

Now consider how you can orient $\vec a_{P,S1}$ and $-\vec a_{P,S2}$ to maximize and minimize the magnitude of $\vec a_{S2,S1}$. Just fix the direction of $\vec a_{P,S1}$ in the x-direction and vary the direction of $-\vec a_{P,S2}$.

Is this right ? I reversed the direction of $\vec a_{P,S2}$, so when $\vec a_{P,S2}$ and $\vec a_{P,S1}$ point at opposite directions, we get maximum $\vec a_{S1,S2}$ and we get minimum(zero) $\vec a_{S1,S2}$ when $\vec a_{P,S2}$ = $\vec a_{P,S1}$ ?

 

[Monsterboy]

Somebody please confirm if the vector diagram is right, I need to make notes. @kuruman you have something to say ?

 

[vela]

Both diagrams are wrong. You made the same mistake earlier in the thread. See posts 5 through 9.

That said, your reasoning about the range of magnitudes of $\vec a_{S1,S2}$ is correct.

I'll add that you really shouldn't need us to confirm the correctness of your answers. You need to construct your understanding so that you can be reasonably confident on your own that an answer is correct, not relying one some authority to tell you that.

 

[kuruman]

The two velocities and the relative velocity, by definition, form a closed triangle. This means that the sides (the two speeds and the relative speed) must obey the triangle inequality.

What I had to say is this:
The statement of the question specifies that the magnitudes of the two accelerations are constant but says nothing about their direction. What about centripetal accelerations?

Imagine a circle of radius 1 m in the lab frame. Frames $S^1$ and $S^2$ are diametrically situated on it and are moving with constant angular speed $\omega=2~\text{rad/s}$. The particle is at rest at the center of the circle.

From the point of view of $S^1$:
1. The particle is undergoing uniform circular motion at radius 1 m and has a constant centripetal acceleration of magnitude $a_c=\omega^2r=4~m/s^2.$
2. 1. Frame $S^2$ is undergoing uniform circular motion at radius 2 m and has constant centripetal acceleration $a_c=\omega^2r=8~m/s^2.$

Clearly, the view from $S^2$ is the same. Now for the choices.

(a) The frames must be at rest with respect to each other.
False. In the lab frame, the frames have equal and opposite velocities relative to each other at all times. This means that they can never be at rest relative to each other.

(b) The frames may be moving with respect to each other but neither should be accelerated with
respect to the other.
False. Each frame has centripetal acceleration in the frame of the other.

(c) The acceleration S2 with respect to S1 may either be zero or $8~m/s^2$.
False. It is only $8~m/s^2$ and never zero.

(d) The acceleration of S2 with respect to S1 may be anything between zero and $8~m/s^2$.
False. It cannot be anything other than $8~m/s^2$.

If the statement of the problem specified that the acceleration of each frame does not change direction, then answer (d) is always True.

I had a private conversation with @haruspex about this. His opinion is that

Option (d) does not preclude exactly 8, regardless of the endpoint subtlety. All it requires the solver to do is to show that for any value from 0 to 8 there is a set-up which achieves that. Since exactly 8 is achievable, it is true whether you read it as including exactly 8 or not. Similarly, exactly zero.​

I guess it depends on how one interprets "may be anything between zero and $8~m/s^2$". It could be taken to mean "all values between zero and $8~m/s^2$ are possible" or "only a single value between zero and $8~m/s^2$ is possible". I favor the former.

My opinion is that for a statement to be True (or False), it must be so in all cases. If there are some physical situations in which the statement is True and others in which it is False, then one cannot say for sure that it is True or False but say instead that it is Sometimes True and Sometimes False (Option 3). Consider, for example, the following two statements
1. An object that moves at constant speed in a straight line does not accelerate.
This statement is always True because it describes constant velocity.
2. An object that moves at constant speed does not accelerate.
This statement is Sometimes True and Sometimes False depending on whether the path is straight or curved.

Thus, in the physical situation of the circling cars that I described here, statement (d) is False. If Option 3 were allowed in this question, then it would be the correct answer because we have physical situations in which choice (d) is True (the accelerations are in straight lines of varying directions) and another physical situation in which choice (d) is False (the accelerations are centripetal.)

It is because of these considerations that I posted #13. I still think the problem could be phrased better.

 

[PeroK]

(d) The acceleration of S2 with respect to S1 may be anything between zero and $8~m/s^2$.
False. It cannot be anything other than $8~m/s^2$.

If the statement of the problem specified that the acceleration of each frame does not change direction, then answer (d) is always True.

I had a private conversation with @haruspex about this. His opinion is that

Option (d) does not preclude exactly 8, regardless of the endpoint subtlety. All it requires the solver to do is to show that for any value from 0 to 8 there is a set-up which achieves that. Since exactly 8 is achievable, it is true whether you read it as including exactly 8 or not. Similarly, exactly zero.​

I agree with @haruspex. More simply, if we have linear acceleration in opposite directions, then the magnitude of the acceleration is precisely $8m/s^2$. That's a specific case. Without knowing further details of the scenario, the answer is d). If we add further information, then some of the range $0-8m/s^2$ may be excluded.

 

[kuruman]

I agree with @haruspex. More simply, if we have linear acceleration in opposite directions, then the magnitude of the acceleration is precisely $8m/s^2$. That's a specific case. Without knowing further details of the scenario, the answer is d). If we add further information, then some of the range $0-8m/s^2$ may be excluded.

I do not disagree either with you or with @haruspex. All I am pointing out is that since the statement of the question does not exclude my example explicitly, choice (d) must be True in that case also. True means True in all scenarios that match the parameters of the problem else the parameters need to be sharpened, in this case by saying that the accelerations are in straight lines that could form any angle between them.

 

[PeroK]

I do not disagree either with you or with @haruspex. All I am pointing out is that since the statement of the question does not exclude my example explicitly, choice (d) must be True in that case also. True means True in all scenarios that match the parameters of the problem else the parameters need to be sharpened, in this case by saying that the accelerations are in straight lines that could form any angle between them.

Sorry, that's absurdist logic. If someone has a British coin, say, then it could be anything from 1p to £2. That's the answer - it could be anyone of those. If you have a 5p piece, you can't declare that to be an exception to the rule - because it's only one of the possibilities and not them all simultaneously.

 

[kuruman]

I don't think I explained my point in a way that can be understood. I will have to rethink how to go about it.

 

[Orodruin]

I don't think I explained my point in a way that can be understood. I will have to rethink how to go about it.

The problem is that you have picked a particular case of the possible cases. Based only on the information in the problem, there are other cases that would satisfy the other values. The question is about what values are possible given the information given, not a particular case.

 

[Orodruin]

then it could be anything from 1p to £2

I’d like a $\pi$ pence coin, please. ;)

 

[Monsterboy]

Both diagrams are wrong. You made the same mistake earlier in the thread. See posts 5 through 9.

That said, your reasoning about the range of magnitudes of $\vec a_{S1,S2}$ is correct.

I'll add that you really shouldn't need us to confirm the correctness of your answers. You need to construct your understanding so that you can be reasonably confident on your own that an answer is correct, not relying one some authority to tell you that.

I haven't studied physics since 2010 and that was just basic stuff, so anything I think is correct is probably not.
Anyway, I hope the below is correct.