# Approximating maximum speed of a particle in terms of angular speed

#### jumbo1985

Here is a rule that governs the motion of particles in my hypothetical Cartesian plane universe:

- a particle can move a maximum distance of Vx units along the x-axis and Vy units along the y-axis per one unit of time.
- the acceleration along the axes is also bounded. The limits are Ax for the x-axis and Ay for the y-axis.

Suppose that a particle moves around a circular path of radius R. Given the maximum per axis velocity/acceleration restrictions above, how can I compute the maximum angular displacement per unit of time?

Any tips greatly appreciated. Thanks

#### romsek

MHF Helper
One problem is that you could have a pulse of acceleration for a vanishingly small duration that would essentially contribute no angular displacement but would nevertheless blow past your acceleration limits.

Are you constrained to constant angular velocity?

#### jumbo1985

Thanks Romsek. No, the angular velocity is not constant.

Lets's assume that the particle is moved by a machine consisting of two robotic arms. One arm only moves "horizontally" (along the x-axis), the other one only moves "vertically" (along the y-axis). There are restrictions on how fast the individual arms travel/accelerate.

What I need to do is translate the component-wise restrictions on the motion into angular terms so that circular motion can be analyzed in simpler terms.

Maximum angular speed/acceleration are two main quantities I seek to derive.

Initial/end speed at the start/end of movement is another quantity I'd like to have represented in terms of the angular equivalents.

I understand that a small pulse of acceleration by one of the arms could have no significance on the displacement but the limits of my theoretical machine must be respected or things blow up (I can set the limits arbitrarily high if I want to).

#### topsquark

Forum Staff
I don't know if anyone can give you a full answer to your questions. For instance are you considering that the larger the linear speed in the x direction the more resistance you run into? Or are you merely constrained by a constant maximum acceleration? There are a lot of dynamics waiting there to be explored depending on the answer.

However we can do a bit. The total acceleration on a rotating object is the vector sum of the centripetal acceleration and the linear (in the direction of motion) acceleration. Centripetal acceleration is in the direction of motion an is equal to $$\displaystyle a_c = \frac{v^2}{r}$$, so your max centripetal acceleration would be at $$\displaystyle a_c = \frac{V_x^2 + V_y^2}{r}$$. The max. linear acceleration is going to depend on the sizes of Ax and Ay, but you can find the max there by taking $$\displaystyle A_{max} = \sqrt{A_x^2 + A_y^2}$$. So the total acceleration here will be $$\displaystyle \sqrt{A_{max}^2 + a_c^2}$$. But it's not quite that easy because this is going to depend on which way the total acceleration vector is pointing due to the constraints on the acceleration components.

Better than that I cannot do without more details.

-Dan

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#### romsek

MHF Helper
ok well the angular velocity limit is pretty straightforward.

You are going to be constrained by the smaller of the two cartesian velocity limits so let $v_m = \min({vx}_m, {vy}_m)$

$R \dot{\phi} \leq v_m$

where $\dot{\phi}$ is the angular velocity

with circular motion all acceleration is purely normal and has magnitude $a_n = \dot{\phi}^2 R$

at values of $\phi$ that are a multiple of $\dfrac \pi 2$ this normal acceleration will be purely in one of the cartesian axis directions and thus

letting $a_m = \min({ax}_m, {ay}_m)$

$a_n = \dot{\phi}^2 R \leq a_m$

combining these you have

$\dot{\phi} \leq \min\left(\dfrac {v_m}{R}, \sqrt{\dfrac{a_m}{R}}\right)$

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• 2 people