Would particle P be under non-unform circular motion?

[vcsharp2003]

Homework Statement

A particle P, as shown in diagram below, is moving at a velocity of $2 \hat i \, {m/s}$ at time $t=0$ under a constant acceleration of $2 \hat i - 2 \hat j \, m/s^2$. Would the particle move in a circular path? Provide explanation.

Relevant Equations

$F_c= \dfrac {mv^2} {r}$
$a_t= \alpha r$

At t= 0, we can see that the particle P has a radial acceleration of $-2\hat j$ and a tangential acceleration of $2 \hat i$. The radial acceleration will tend to move it in a circle of a certain radius, whereas the tangential acceleration will tend to displace it parallel to x- axis.

Consider a very small interval of time after the time t = 0. At the end of this very small interval of time, the vector sum of above two displacements is not going to be on a circle as shown in diagram below. In this diagram, the displacements due to tangential and radial accelerations are shown as $d_2$ and $d_1$ respectively after a very small time interval from t = 0. We can see that net displacement is going to place the particle slightly above the circular path and therefore, the particle will not move along a circular path.

 

[haruspex]

It doesn’t specify a circle about the origin.

 

[vcsharp2003]

It doesn’t specify a circle about the origin.

In my diagram, O is the center of a circle that would be traversed if there was no tangential acceleration. So, O is not the origin of the x-y axes system.

 

[archaic]

if the force is such that it would make the particle move in a circular path, then it should verify$$|\mathbf F|=\frac{m|\mathbf v(t)|^2}{R}$$suppose that you already know $m$ and that there is such an $R$. you are given $\mathbf F(=m\mathbf a)$ and you can deduce $\mathbf v(t)$ from the given data.
is that equation verified?

 

[haruspex]

In my diagram, O is the center of a circle that would be traversed if there was no tangential acceleration. So, O is not the origin of the x-y axes system.

Then I do not understand your diagram.
You define $\vec d_1, \vec d_2$ as radial and tangential accelerations about some presumed centre, but you do not show them as perpendicular.

The constant radial acceleration

It is given as a constant acceleration, not as a constant radial acceleration.

if the force is such that it would make the particle move in a circular path, then it should verify$$|\mathbf F|=\frac{m|\mathbf v(t)|^2}{R}$$

Aren’t you assuming uniform circular motion?

 

[archaic]

Aren’t you assuming uniform circular motion?

even in non-uniform circular motion, $$|\mathbf F_c|=|\sum\mathbf F|=\frac{m|\mathbf v|^2}{R}$$right?
the idea is to do this by contradiction. there will be a constant term equal to a varying one.

 

[vcsharp2003]

You define d→1,d→2 as radial and tangential accelerations

These are displacements due to tangential and radial accelerations.

 

[vcsharp2003]

It is given as a constant acceleration, not as a constant radial acceleration.

Yes, that is true and my mistake.

But since I am considering a very small interval of time, I am showing the full circle along which a very small arc of particle's trajectory would lie due to acceleration perpendicular to its velocity at time t= 0.

 

[vcsharp2003]

It is given as a constant acceleration, not as a constant radial acceleration.

I have corrected this in my original post.

 

[haruspex]

Yes, that is true and my mistake.

But since I am considering a very small interval of time, I am showing the full circle along which a very small arc of particle's trajectory would lie due to acceleration perpendicular to its velocity at time t= 0.

One difficulty with supposing a small interval of time is that you are starting with accelerations, so one level of such integration gives velocity change, not displacement.
Another is that you need to be careful to ignore second order small differences, or you may wrongly conclude the path cannot be circular.

I think there is a much easier way to answer the question. If the acceleration is constant, what can you say about its distance from some given point over the long term?

 

[vcsharp2003]

I think there is a much easier way to answer the question. If the acceleration is constant, what can you say about its distance from some given point over the long term?

You mean get $\vec {v(t)}$ by integrating $\dfrac {d {\vec v}} {dt} = 2 \hat i -2 \hat j$ and then get position vector at a time t $\vec {s(t)}$ by integrating the velocity equation.

 

[vcsharp2003]

If the acceleration is constant, what can you say about its distance from some given point over the long term?

If it's one dimensional motion, then constant acceleration would cause distance from a given point to increase over a long term. But, I am not sure if the same occurs in two dimensional motion.

 

[Doc Al]

I have corrected this in my original post.

As far as I can see, your original post still shows a constant acceleration. What should it show?

 

[vcsharp2003]

If the acceleration is constant, what can you say about its distance from some given point over the long term?

In this 2 dimensional motion scenario, we could consider the acceleration component along x-axis separately from acceleration component along y axis, since for any vector we can consider it's components instead of the original vector.

x axis: The acceleration is + 2 m/s².

y axis: The acceleration is -2 m/s².

So, we can say that distance from its position at t=0 is going to increase as time increases. So, if we take an arbitrary point in x-y plane, then after some very large interval of time the distance of particle P would be increasing from this arbitrary point.

 

[vcsharp2003]

As far as I can see, your original post still shows a constant acceleration. What should it show?

The acceleration of particle P is constant, but saying that radial acceleration is constant is incorrect. I corrected the parts where it said constant radial acceleration.

 

[Doc Al]

The acceleration of particle P is constant, but saying that radial acceleration is constant is incorrect. I corrected the parts where it said constant radial acceleration.

OK. Why would you think that a constant acceleration would produce a circular trajectory? (Compare to a case that I suspect you are familiar with: a particle under the influence of gravity.)

 

[vcsharp2003]

OK. Why would you think that a constant acceleration would produce a circular trajectory? (Compare to a case that I suspect you are familiar with: a particle under the influence of gravity.)

Yes, you are correct. There is no reason to consider it like circular motion. It's a case of 2D motion with constant acceleration.

It is very similar to projectile motion under gravity except that it has a horizontal acceleration and its vertical acceleration is -2 m/s².

 

[Doc Al]

It is very similar to projectile motion under gravity except that it has a horizontal acceleration and its vertical acceleration is -2 m/s².

Exactly. And to see things more clearly, feel free to assign a new set of axes so that the acceleration is purely "vertical".

 

[vcsharp2003]

Exactly. And to see things more clearly, feel free to assign a new set of axes so that the acceleration is purely "vertical".

Acceleration cannot be just vertical since there is a horizontal as well as a vertical component relative to ground.

 

[Doc Al]

Acceleration cannot be just vertical since there is a horizontal as well as a vertical component relative to ground.

Define a new set of axes (x', y') that are at an angle with respect to x,y so that the acceleration is in the y' direction. Then the trajectory in that coordinate system will be familiar.

 

[vcsharp2003]

Define a new set of axes (x', y') that are at an angle with respect to x,y so that the acceleration is in the y' direction. Then the trajectory in that coordinate system will be familiar.

That will require a rotated axes system.

 

[Doc Al]

That will require a rotated axes system.

Sure.

 

[vcsharp2003]

Define a new set of axes (x', y') that are at an angle with respect to x,y so that the acceleration is in the y' direction. Then the trajectory in that coordinate system will be familiar.

I have used a new rotated system of axes x'-y', and then there is no acceleration along x' axis and a constant acceleration in negative y' axis direction. The case then becomes like a projectile motion under gravity with particle thrown down at 45^o to vertical. We would then end up with a perfect parabolic path in the x'-y' axes system.

Therefore, the trajectory of the particle will not be a circular path in the x-y axes system. The distance of particle P from its postion at t= 0 will keep on increasing with time due to its parabolic trajectory.

Actually, any two dimensional motion with constant acceleration and an initial velocity that is not parallel to constant acceleration, can be analyzed in this manner and will always follow a parabolic path in a rotated axes system.

 

[Doc Al]

We would then end up with a perfect parabolic path in the x'-y' axes system.

Therefore, the trajectory of the particle will not be circular path in x-y axes system.

Good! Switching to the x'-y' axes allows you to immediately recognize the trajectory as a parabola.

 

[Steve4Physics]

Homework Statement:: A particle P,...is moving ... under a constant acceleration of $2 \hat i - 2 \hat j \, m/s^2$. Would the particle move in a circular path? Provide explanation.

Can I offer a simpler way to answer the original Homework Statement?

The direction of acceleration during circular motion is constantly changing.
$2 \hat i - 2 \hat j \, m/s^2$ has a fixed direction.
From which the answer follows.

(It is interesting to then ask: what form does an acceleration vector need to have to produce circular motion?)

 

[vcsharp2003]

It is interesting to then ask: what form does an acceleration vector need to have to produce circular motion?

The acceleration vector needs to be always perpendicular to velocity of particle for circular motion to occur i.e. its direction is always changing as the particle moves along a circle.

 

[Doc Al]

The acceleration vector needs to be always perpendicular to velocity of particle for circular motion to occur

Careful. For uniform circular motion ,the acceleration vector is always perpendicular to the velocity. But for non-uniform circular motion the acceleration will have a component parallel to the velocity (often called the tangential component). But you are correct that there must be a component perpendicular to the velocity (the radial component) for circular motion to occur.

But the key point, as made by @Steve4Physics, is that for any kind of circular motion the acceleration vector must be changing direction as the particle moves along that circle. Since the given acceleration in this problem has a fixed direction, what can you immediately conclude?

 

[vcsharp2003]

Careful. For uniform circular motion ,the acceleration vector is always perpendicular to the velocity. But for non-uniform circular motion the acceleration will have a component parallel to the velocity (often called the tangential component). But you are correct that there must be a component perpendicular to the velocity (the radial component) for circular motion to occur.

But the key point, as made by @Steve4Physics, is that for any kind of circular motion the acceleration vector must be changing direction as the particle moves along that circle. Since the given acceleration in this problem has a fixed direction, what can you immediately conclude?

The acceleration vector is constant that points in a south-east direction always. But, acceleration does have a perpendicular component to initial velocity pointing in a south direction. This component is always pointing south at all instants of time no matter what the velocity direction is and hence it cannot produce circular motion.

 

[Doc Al]

Good. Circular motion (whether uniform or not) requires the acceleration to change direction. Since the given acceleration remains fixed, it cannot possibly describe circular motion.

That's all you need to answer the question. But you can also deduce rather easily that the given acceleration produces a parabolic trajectory.

 

[vcsharp2003]

But you can also deduce rather easily that the given acceleration produces a parabolic trajectory.

Yes that was an excellent and beautiful solution. The parabolic path is a rotated parabola.