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Problem

Say I have the following equation.

$y=f(\theta)$

Where

$\theta = \int\int\alpha$

Is it possible to express the equation in terms of $\alpha$ and not the double integral of it?

Origin

This problem originates from the problem shown in the image below.

Problem.jpg

Where the equations $F_y=M_ysin(\frac{\pi}{2}-\theta)$ and $F_x=M_ycos(\frac{\pi}{2}-\theta)$ both depend on the angle $\theta$ but where only the angular acceleration $\alpha$ is known.

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  • $\begingroup$ Very confusing question. Could you type in the original problem? Is it a pendulum? Is the force constant? What is exactly the question? $\endgroup$ – Andrei Apr 8 at 18:21
  • $\begingroup$ @Andrei No, it's not confusing. How can $F_y=M_ysin(\frac{\pi}{2}-\theta)$ and the other equation be expressed in terms of $\alpha$ and not $\theta$. Simply put, remove the double integral. $\endgroup$ – Full_Nitrous Apr 8 at 20:43
  • $\begingroup$ I meant that the problem in the image makes no sense to me. You have some formulas there, and some pictures, but I do not know what they represent $\endgroup$ – Andrei Apr 8 at 20:45
  • $\begingroup$ @Andrei Oh, yes ok. The force $F$ is the force of a gimballed rocket engine with the angle $v$ mounted on a rocket body which has a orientation angle of $\theta$. The forces $F_y$ and $F_x$ are the translative forces resulting from the force $F$ disregarding that the rotation itself creates a force. $\endgroup$ – Full_Nitrous Apr 8 at 20:47
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With the additional comments, the angular acceleration $\alpha$ is a constant (not depending of $\theta$). You can then write $$\theta=\theta_0+\omega_0t+\frac 12\alpha t^2$$ Here $\theta_0$ is the original angle of the rocket, and $\omega_0$ is the original angular velocity. $t$ is time.

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  • $\begingroup$ Using $t^2$ means the same as the integral. This does not solve the problem. $\endgroup$ – Full_Nitrous Apr 8 at 21:07
  • $\begingroup$ Then the answer is no. Since $\theta$ depends on the time dependent value of $\alpha$, there is no other way to express $\theta$. $\endgroup$ – Andrei Apr 8 at 21:10
  • $\begingroup$ The question is not how to express $\theta$ rather how to change the entire equation so that $\alpha$ can be used instead. $\endgroup$ – Full_Nitrous Apr 8 at 21:11
  • $\begingroup$ Ok, so here, put more generally, what are the equations of motion for the rocket spesified? Preferrably the same rocket but in three dimensions with two gimbal axis. Though the solution should be easy to implement in 3D. Should i re-ask the question asking for the equations of motion instead? $\endgroup$ – Full_Nitrous Apr 8 at 21:13
  • $\begingroup$ Probably a good idea, but let me just try to clarify it now. For the equations of motion (in the differential form) you will write $\alpha$ as a function of $\theta$. That is $$\alpha=\ddot\theta=f(\theta)$$ $\endgroup$ – Andrei Apr 8 at 21:21

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