Trajectory of a turning particle

In summary, the problem involves finding the trajectory of a particle with a constant speed and increasing angular acceleration, resulting in a spiral trajectory. The particle has an initial position and velocity vector, and the velocity vector is rotated at a rate determined by the angular acceleration. Differentiating the equation for constant speed and using the fact that lateral acceleration is orthogonal to the plane of motion, we can write an equation for the derivative of the velocity vector in terms of the velocity vector and a unit vector orthogonal to the plane of motion. This can then be used to integrate and find an expression for the position vector as a function of time.
  • #1
rovim
6
0
In this problem, I need to find the trajectory of a particle (as a function of time) which moves at a speed 's' but also turns at an increasing rate; angular acceleration α. The trajectory looks like a spiral which converges to a point.

The particle has an initial position vector p and a velocity vector v. So without an angular velocity or 'turning effect' the particle should simply trace out the path "p(t) = ∫ v dt", however the particle has an angular acceleration which means that the velocity vector is rotated at the rate α, therefore vnew = R(Δθ) vold (R(θ) is the rotation matrix to rotate a 2d vector θ degrees).

I'm looking at this problem in a 2d sense.

An example of this would be a car that is traveling a constant speed but the driver starts turning the wheel at a constant rate so that the car turns sharper and sharper.

I've tried tackling this problem by expressing the velocity vector and angular velocity as a complex numbers and integrating their product, as well as trying a method using the rotation matrix, but on both occasions I'm stuck with 'dt's' inside trigonometric functions while integrating.
 
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  • #2
rovim said:
In this problem, I need to find the trajectory of a particle (as a function of time) which moves at a speed 's' but also turns at an increasing rate; angular acceleration α. The trajectory looks like a spiral which converges to a point.

The particle has an initial position vector p and a velocity vector v. So without an angular velocity or 'turning effect' the particle should simply trace out the path "p(t) = ∫ v dt", however the particle has an angular acceleration which means that the velocity vector is rotated at the rate α, therefore vnew = R(Δθ) vold (R(θ) is the rotation matrix to rotate a 2d vector θ degrees).

I'm looking at this problem in a 2d sense.

An example of this would be a car that is traveling a constant speed but the driver starts turning the wheel at a constant rate so that the car turns sharper and sharper.

I've tried tackling this problem by expressing the velocity vector and angular velocity as a complex numbers and integrating their product, as well as trying a method using the rotation matrix, but on both occasions I'm stuck with 'dt's' inside trigonometric functions while integrating.
Let ##\hat k## be the unit vector orthogonal to the plane of motion. Can you write a vector equation that relates ##\dot {\vec v}## to ##\vec v##, ##\hat k## and some scalar function ##\alpha(t)##? Done correctly, it will entail the information that the speed is constant.
 
  • #3
haruspex said:
Let ##\hat k## be the unit vector orthogonal to the plane of motion. Can you write a vector equation that relates ##\dot {\vec v}## to ##\vec v##, ##\hat k## and some scalar function ##\alpha(t)##? Done correctly, it will entail the information that the speed is constant.
Thanks for responding, but I don't really grasp your idea with my limited knowledge of vector calculus. Wouldn't adding a perpendicular component cause the speed to grow unboundedly? To me, it doesn't seem to be a true rotation. I might be completely off track, so if someone could provide me with a semi-detailed derivation or point to a similar problem, that would be a life saver.
 
  • #4
rovim said:
Thanks for responding, but I don't really grasp your idea with my limited knowledge of vector calculus. Wouldn't adding a perpendicular component cause the speed to grow unboundedly? To me, it doesn't seem to be a true rotation. I might be completely off track, so if someone could provide me with a semi-detailed derivation or point to a similar problem, that would be a life saver.
If you want the particle to turn there has to be a perpendicular component. If you want the speed constant, there can only be a perpendicular acceleration. If it is a constant magnitude acceleration the particle will move in a circle, but you wanted a spiral, so it has to vary. (I did not suggest it should increase without limit.)
 
  • #5
haruspex said:
If you want the particle to turn there has to be a perpendicular component. If you want the speed constant, there can only be a perpendicular acceleration. If it is a constant magnitude acceleration the particle will move in a circle, but you wanted a spiral, so it has to vary. (I did not suggest it should increase without limit.)
Ah yes I understand that now. With some working out I end up with the equation:
Δp/Δt = s##\hat v## + ##\frac{1}{2}## srt2##\hat k ##

v is the direction of the velocity, s is the speed of the movement and r is the angular acceleration.

Now, I have absolutely no idea to integrate this to get an expression for p.
 
  • #6
rovim said:
Ah yes I understand that now. With some working out I end up with the equation:
Δp/Δt = s##\hat v## + ##\frac{1}{2}## srt2##\hat k ##

v is the direction of the velocity, s is the speed of the movement and r is the angular acceleration.

Now, I have absolutely no idea to integrate this to get an expression for p.
I defined ##\hat k ## as orthogonal to the plane of motion. The lateral acceleration is therefore orthogonal to that. Since you want constant speed, we have ##|\vec v|^2=\vec v.\vec v## constant. What do you get if you differentiate that that? Can you then see how to write ##\dot {\vec v}## in terms of ##\vec v## and ##\hat k## ?
 
  • #7
haruspex said:
I defined ##\hat k ## as orthogonal to the plane of motion. The lateral acceleration is therefore orthogonal to that. Since you want constant speed, we have ##|\vec v|^2=\vec v.\vec v## constant. What do you get if you differentiate that that? Can you then see how to write ##\dot {\vec v}## in terms of ##\vec v## and ##\hat k## ?
Hmmm. Differentiating ##|\vec v|^2=\vec v.\vec v## only seems to prove the result ##\dot {\vec v} .\vec v = 0## which gives me the idea that I should completely replace the ##\hat k## component with ##\dot {\vec v}##. But still I haven't been able find the magnitude.

I believe I'm lacking some essential knowledge in vector calculus to be able to solve this. So putting this as mathematically as I can, the problem describes 4 equations I really don't know how to put together to find p(t):
##\vec p (t) = \int_0^t \! \vec v (t) \, \mathrm{d}t + p_{initial}##
##\vec v (t) = \int_0^t \! \ \dot{\vec v} (t) \, \mathrm{d}t + v_{initial}##
##\dot {\vec v} \cdot \vec v = 0##
##| \dot {\vec v} (t) | = \alpha (t) | \vec v (t) | ##
 
  • #8
rovim said:
Hmmm. Differentiating ##|\vec v|^2=\vec v.\vec v## only seems to prove the result ##\dot {\vec v} .\vec v = 0## which gives me the idea that I should completely replace the ##\hat k## component with ##\dot {\vec v}##. But still I haven't been able find the magnitude.

I believe I'm lacking some essential knowledge in vector calculus to be able to solve this. So putting this as mathematically as I can, the problem describes 4 equations I really don't know how to put together to find p(t):
##\vec p (t) = \int_0^t \! \vec v (t) \, \mathrm{d}t + p_{initial}##
##\vec v (t) = \int_0^t \! \ \dot{\vec v} (t) \, \mathrm{d}t + v_{initial}##
##\dot {\vec v} \cdot \vec v = 0##
##| \dot {\vec v} (t) | = \alpha (t) | \vec v (t) | ##
##\dot {\vec v} .\vec v = 0## tells you that those two vectors are orthogonal. ##\dot {\vec v} ## is also orthogonal to ##\hat k##. If a vector is known to be orthogonal to two others, how can you write it as a scalar multiple of a certain function of those two vectors?
 
  • #9
haruspex said:
##\dot {\vec v} .\vec v = 0## tells you that those two vectors are orthogonal. ##\dot {\vec v} ## is also orthogonal to ##\hat k##. If a vector is known to be orthogonal to two others, how can you write it as a scalar multiple of a certain function of those two vectors?
So ##\vec v = \alpha (t) \vec v \times \hat k ##?
 
  • #10
It's been a few days and I still have no idea. Could anyone atleast show me what topics to look into to solve this problem?
 

1. What is the trajectory of a turning particle?

The trajectory of a turning particle refers to the path that a particle takes as it rotates around a fixed point or axis. This path can be either circular or elliptical, depending on the forces acting on the particle.

2. What factors affect the trajectory of a turning particle?

The trajectory of a turning particle is affected by several factors, including the initial velocity and direction of the particle, the magnitude and direction of the force acting on the particle, and any external forces such as friction or air resistance.

3. How can the trajectory of a turning particle be calculated?

The trajectory of a turning particle can be calculated using mathematical equations and principles, such as the laws of motion and rotational dynamics. These equations take into account the various forces acting on the particle and can predict its path over time.

4. What is the difference between a circular and elliptical trajectory?

A circular trajectory is a path in which the particle maintains a constant distance from a fixed point, while an elliptical trajectory is a path in which the distance from the fixed point varies. This variation is due to the changing magnitude and direction of the forces acting on the particle.

5. How does angular momentum affect the trajectory of a turning particle?

Angular momentum, which is a measure of an object's rotational motion, plays a crucial role in determining the trajectory of a turning particle. The conservation of angular momentum states that the total angular momentum of a system remains constant, so any changes in direction or speed of the particle will affect its trajectory.

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