Maximum tangential velocity from maximum linear velocities

In summary, the problem presents a particle moving along a circular arc centered at the origin in a 2D plane. The angular displacement function for the time intervals T0 through T3 is given, along with acceleration and velocity constraints in Cartesian coordinates. The goal is to calculate the maximum values of A_T and V_T in terms of linear quantities A_max and V_max, which put a limit on A_T and V_T. The problem also involves calculating the time intervals T1, T2, and T3. The approach involves breaking down the problem into components and using numerical simulation or plotting to gain insight on the extrema and dependencies on time. However, the unknown values of T1, T2, and T3 make it difficult to
  • #1
jumbo1985
19
1
I have a particle that moves along a circular arc centered at origin in 2D plane.

I have the following angular displacement function for time T0 through T3 and the following acceleration and velocity constraints in Cartesian coordinates. At t = T0, theta = 0, velocity = 0. At t = T3, velocity = 0.

I want to calculate the maximum A_T and V_T in terms of the linear A_max and V_max quantities (A_max and V_max put a limit on A_T and V_T). I want to eventually be able to calculate T1, T2, T3.

[tex]
\forall t \in [T_0, T_1]
\begin{cases}
\frac{d^2\theta}{dt^2}(t) = A_T \\
\frac{d\theta}{dt}(t) = \dot{\theta}_{0} + A_T(t-T_0) \\
\theta(t) = \theta_{0} + \dot{\theta}_{0}(t-T_0)+\frac{1}{2}A_T(t-T_0)^2 \\
\end{cases}

\\

\forall t \in [T_1, T_2]
\begin{cases}
\frac{d^2\theta}{dt^2}(t) = 0 \\
\frac{d\theta}{dt}(t) = V_T \\
\theta(t) = \theta(T_1)+V_T(t-T_1) \\
\end{cases}

\\

\forall t \in [T_2, T_3]
\begin{cases}
\frac{d^2\theta}{dt^2}(t) = -A_T \\
\frac{d\theta}{dt}(t) = \dot{\theta}(T_2) - A_T(t-T_2) \\
\theta(t) = \theta(T_2) + \dot{\theta}(T_2)(t-T_2) - \frac{1}{2}A_T(t-T_2)^2 \\
\end{cases}
\\
\\
\vec{A_{max}} = \begin{pmatrix}
A_{max,x} \\
A_{max,y} \\
\end{pmatrix}
\\
\vec{V_{max}} = \begin{pmatrix}
V_{max,x} \\
V_{max,y} \\
\end{pmatrix}
\\
\vec{Position} =
\begin{pmatrix}
x \\
y \\
\end{pmatrix} =
\begin{pmatrix}
\cos(\theta(t)) \\
\sin(\theta(t)) \\
\end{pmatrix}
[/tex]

At first I figured I could solve
[tex]\vec A =
\begin{pmatrix}
\frac{d^2}{dt}[\cos(\theta(t))] \\
\frac{d^2}{dt}[\sin(\theta(t))]\\
\end{pmatrix}
[/tex]

for A_T at t = T0 with all of the initial conditions applied but I'm afraid this is not correct.

How do I approach this problem?

Thanks
 
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  • #2
Your sequence is: constant angular acceleration in ##[T_0, T_1]##, uniform angular motion in ##[T_1, T_2]## and constant angular deceleration in ##[T_2, T_3]## with the same angular acceleration, but opposite sign..
You don't mention the radius ##r## of the circle.
jumbo1985 said:
I want to calculate the maximum A_T and V_T in terms of the linear A_max and V_max quantities (A_max and V_max put a limit on A_T and V_T).
Your expressions provide these numbers: maximum angular velocity ##A_T (T_1-T_0) ## at ##T_1##. Linear acceleration at that point (value unknown if ##r## is unknown) and linear velocity (idem) both at maximum.
etc etc.
 
  • #3
Yes, sorry. I wanted to make the radius R be variable and I forgot to include it both the description and the equations.

I think I should have stated my question better. I'm looking for the largest magnitudes of A_T and V_T such that
[tex]|A_x| =|\frac{d^2}{dt^2}(R\cos(\theta(t)))| \leq A_{max,x}\\
|A_y| =|\frac{d^2}{dt^2}(R\sin(\theta(t)))| \leq A_{max,y}\\
|V_x| =|\frac{d}{dt}(R\cos(\theta(t)))| \leq V_{max,x}\\
|V_y| =|\frac{d}{dt}(R\sin(\theta(t)))| \leq V_{max,y}\\[/tex]

If I manage to come up with something, I'll update the post.
Thanks
 
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  • #4
Well, you have an expression for ##\theta(t)## so you can differentiate it.
 
  • #5
For
[tex]t \in [T_0,T_1], \dot{\theta}(T_0) = 0, \theta(T_0) = 0, R > 0, A_{max,x} > 0[/tex]
Analyzing A_x. After a bit of simplification, I arrive at
[tex]|(-1)(A_T\sin(\frac{1}{2}A_T(t-T_0)^2)+A_T^2(t-T_0)^2\cos(\frac{1}{2}A_T(t-T_0)^2))| \leq \frac{|A_{max,x}|}{R} = \frac{A_{max,x}}{R}[/tex]
I'm not quite sure how to get an upper bound for A_T from this.

I would then like to perform the same analysis for
[tex]t \in [T1,T2], t \in [T2,T3][/tex]
and combine the estimates. I'm not sure if this approach will work
 
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  • #6
Is ##A_{max,x}## different from ##A_{max,y}## ? Is seems to me that this treatment in cartesian coordinates makes things unnecessarily complicated; that's why I ask.
Can you show your differentiation steps ? Doesn't look right dimension-wise !

If I do ##{d^2\over dt^2 } \left ( R\cos\theta\left (t \right )\right)## I get (with a shorthand dot for time derivative): $$
{d\over dt } \left ( R\cos\theta\left (t \right )\right) = -R\sin\theta\; \dot \theta \Rightarrow $$ $$
{d^2\over dt^2 } \left ( R\cos\theta\left (t \right )\right) = - R\cos\theta\; \dot \theta^2 - R\sin\theta\; \ddot \theta$$with ## \ddot \theta = A_T## and ## \dot \theta = A_T(t-T_0)##
 
  • #7
Yes, the maximum acc/vel magnitudes in cartesian coordinates are not ideal to work with. This is how the world (2d plane) in which my particle moves is described unfortunately. [tex]A_{max,x}[/tex] could be different from [tex]A_{max,y}[/tex] which is why I feel I need to break this down into components.

For [tex]t\in[T_0,T_1][/tex]
I get
[tex]A_x = -R((A_T(t-T_0))^2\cos(0.5A_T(t-T_0)^2) + A_T\sin(0.5A_T(t-T_0)^2))[/tex]

I expect to find A_T in terms of A_x then in terms of A_y and settle for the smaller of the values so that neither A_max,x nor A_max,y are ever exceeded as the particle travels around the circular arc.
 
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  • #8
Our ##A_x## agree. So (in a mixed notation) ##A_x = - \omega^2 x - \alpha y\quad ## with ##(\alpha = A_T)##. and so on.
In uniform circular motion the amplitudes of ##A_x## and ##A_y## are equal, but in uniformly accelerated motion both grow at the same rate, but individual extrema occur at different moments, so the value of ##T_1## comes into play.
jumbo1985 said:
I feel I need to break this down into components
It depends: if ##T_1## is large and ##A_T## is not, then during one revolution the extrema won't differ much; that's when you can simplify and use ##\min (A_x, A_y)## as constraint.
But if you accelerate strongly and change to coasting quickly it's different.

Comparable reasoning for ##T_2, T_3##.

Numerical simulation (or simply drawing ##x(t)## and ##y(t)## from ##T_0## to ##T_3##) might give easy insight
 
  • #9
Thanks for taking the time to respond BvU.
I plotted the x''(t) and y''(t) on the domain from T0 to T1 with some arbitrary numbers and that gave me some interesting insight. I see the amplitudes peaking at time T1. In general I see the dependence on time.

The problem I'm facing is that only T0 is known, theta at T0 and T3 is known but T1, T2, T3 are not known. I'm not sure how to calculate these without knowing A_T and V_T.
 
  • #10
I understand. Is it clear that the number of revolutions during the cycles plays a role in estimating whether the difference between ##A_x{\rm,\; max}## and ##A_y{\rm,\; max}## is relatively unimportant ? (in particular during the stage ##[T_1, T_2]## ?

Can you formulate the optimization problem (apart from what's already in post #1, I mean):
 
  • #11
I should have mentioned that:
[tex]|\theta_{T_3} - \theta_{T_0}| \leq 2\pi[/tex]

ie. the particle will never make more than one full revolution around the origin (both thetas are known).

The particle needs to reach the coasting stage or transition straight from the acceleration to deceleration stage as quickly as possible.

I need to think about everything that was posted so far again from the beginning.
 
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  • #12
jumbo1985 said:
I should have mentioned
Yes. Helps understand why x and y can be different :smile:
jumbo1985 said:
both thetas are known
that too ! The constraint I've been fishing for ...
jumbo1985 said:
The particle needs to reach the coasting stage or transition straight from the acceleration to deceleration stage as quickly as possible.
Perhaps you can consider only Ay in first order if theta(0) = 0 ?
 
  • #13
At [tex]\theta(0)=0 \implies A_y = RA_T[/tex]

Sorry if I misunderstood the question
 

FAQ: Maximum tangential velocity from maximum linear velocities

What is the maximum tangential velocity and how is it related to maximum linear velocities?

The maximum tangential velocity is the highest velocity that an object can achieve while moving in a circular path, and it is directly related to the maximum linear velocities through the radius of the circle. The larger the radius, the higher the maximum tangential velocity can be.

Why is it important to calculate the maximum tangential velocity?

Calculating the maximum tangential velocity is important in many fields of science, such as physics and engineering, as it helps determine the maximum possible speed for an object in circular motion. This information is crucial for designing and predicting the behavior of various systems, such as vehicles and machines.

How do you calculate the maximum tangential velocity from maximum linear velocities?

The formula for calculating the maximum tangential velocity is Vt = ωr, where Vt is the tangential velocity, ω is the angular velocity (which can be calculated from the maximum linear velocity), and r is the radius of the circular path. By plugging in the values for ω and r, you can calculate the maximum tangential velocity.

What factors can affect the maximum tangential velocity?

The maximum tangential velocity can be affected by various factors, such as the radius of the circle, the mass of the object, and any external forces acting on the object. Friction and air resistance can also affect the maximum tangential velocity.

Can the maximum tangential velocity ever be greater than the maximum linear velocity?

No, the maximum tangential velocity can never be greater than the maximum linear velocity. This is because the tangential velocity is dependent on the angular velocity, which is always less than or equal to the linear velocity. Therefore, the maximum tangential velocity will always be less than or equal to the maximum linear velocity.

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