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Sorry for asking this simple question, but really I couldn't find a good document discuss what I need exactly.

I am implementing a flight simulation, but my question is related to physics rather than aerodynamics so I find to ask the question to physics experts.

Suppose that I am having a cuboid (Simple form of the plane) with the following dimension: Length: 14.8m
Height: 4.8m
Depth: 10.0m

The coordinate system is X is right, y is Up and z depth (inside the paper).

I applied a torque on the Y axis, the rectangle begins to gain angular velocity and it rotates in the XZ plane around its center of gravity.

Every thing works fine for now, but after I remove the torque the cuboid should stops i.e. the angular velocity should be decreased till reaches zero. How this happens?

I think this should be due to the moment of inertia, as I am using 3D coordinate system the inertia should be inertia tensor and the right way to calculate the inertia tensor from these dimensions.

If what I thought is right so I need the equation for how the inertia tensor is affecting the angular velocity till the angular velocity reaches zero.

If I am not right, what is the force that affect the cuboid to stop rotating?

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    $\begingroup$ In one word: friction. $\endgroup$ – hdhondt Aug 19 '20 at 9:52
  • $\begingroup$ So what should be the equation, how to calculate it? $\endgroup$ – Ahmed Elhamy Aug 19 '20 at 9:55
  • $\begingroup$ @AhmedElhamy You need to incorporate drag (air resistance) into your equations of motion. This is complicated because the amount and direction of drag will depend on the shape of your plane, how it is moving, and how fast it is moving. This is why aerodynamics is difficult and designing safe control systems for aircraft is even more difficult. $\endgroup$ – gandalf61 Aug 19 '20 at 10:08
  • $\begingroup$ Yes I know the drag equations and how it can be calculated, but drag is not applied on rotation motion only on linear movement. (This is what I have read) $\endgroup$ – Ahmed Elhamy Aug 19 '20 at 10:10
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    $\begingroup$ @AhmedElhamy Then what you have read is wrong, this is known as "viscous torque" if you want to search it up. $\endgroup$ – Charlie Aug 19 '20 at 10:17
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Yes I know the drag equations and how it can be calculated, but drag is not applied on rotation motion only on linear movement. (This is what I have read)

No, it probably probably isn't. What IS true is that most textbooks deal with viscous forces due to linear translation and are silent about rotational viscous drag.

But rotating bodies experience viscous drag too. That's because any element on a rotating body also experiences tangential translational motion.

For simple translational drag, the drag force is given by:

$$F_D=\frac12 \rho v^2 C_D A\tag{1}$$

Now consider the simplest case of a bar rotating around one of its ends $O$:

Viscous rotation

An element $\text{d}x$ at distance $x$ from $O$ has a tangential velocity of:

$$v(x)=\omega x\tag{2}$$ where $\omega$ is the angular velocity about $O$. With $(1)$ we get the infinitesimal drag force $\text{d}F_D$

$$\text{d}F_D=\frac12 \rho v(x)^2 C_D \text{d}A$$

$$\text{d}A=\mu \text{d}x$$

for a uniform bar $\mu=\text{constant}$. $$\text{d}F_D=\frac12 \rho (\omega x)^2 C_D\mu \text{d}x$$ with $(2)$: $$\text{d}F_D=\frac12 \rho \mu \omega^2 C_D x^2 \text{d}x$$ We find the total drag force $F_D$ by simple integration:

$$F_D=\int_0^L\text{d}F_D=\int_0^L\frac12 \rho \mu \omega^2 C_D x^2 \text{d}x$$ $$F_D=\frac12 \rho \mu \omega^2 C_D\int_0^Lx^2 \text{d}x$$ $$F_D=\frac16 \rho \mu \omega^2 C_DL^3$$ where $L$ is the total length.

We can also calculate the total viscous torque $\tau$ from:

$$\text{d}\tau=x\text{d}F_D$$

I'll leave the simple integration to you.

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  • $\begingroup$ Thanks, this is exactly what I need. But I need 2 clarifications: (1) is μ is the inertia? or what it should be?. (2) Can I use the CD that I use in the linear motion for your equation? $\endgroup$ – Ahmed Elhamy Aug 23 '20 at 7:48
  • $\begingroup$ You're welcome! 1). Let's say that the uniform bar has a total area $A$, exposed perpendicularly to the motion and has length $L$, then $\mu=\frac{A}{L}$. 2. Yes, use $C_D$ that is used for linear motion. $\endgroup$ – Gert Aug 23 '20 at 12:41
  • $\begingroup$ Thanks a lot :) $\endgroup$ – Ahmed Elhamy Aug 23 '20 at 12:44
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for your flight simulator you can applied braking torque and then stop the simulation when the angular velocity is zero.

your equation

$$I_y\ddot\varphi(t)=\tau_m(t)+\tau_b(t)$$

where $I_y$ is the inertia about the y axes and $\tau_m$ is the applied torque to accelerate the cuboid and $\tau_b$ to decelerate the cuboid

Simulation

$\tau(t)=\tau_m(t)+\tau_b(t)$ enter image description here

Angular velocity $\dot\varphi$

enter image description here

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  • $\begingroup$ Thanks this is a good simulation, I will try it if I fail to implement @Gert solution. $\endgroup$ – Ahmed Elhamy Aug 23 '20 at 7:50
  • $\begingroup$ I can write you the simulation program $\endgroup$ – Eli Aug 23 '20 at 14:23
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The answer to your question is that in real life any time an object moves in air there are surface forces developing due to the boundary layer of air.

The aerodynamics of rotating objects are very complex (see the magnus effect for example), but the end result is that there is net torque applied opposing the rotational motion, as well as translational forces (lift/drag etc) due to the motion.

Consider a rotating bar, and resolve the velocity $\vec{v} = \vec{\omega} \times \vec{r}$ of the object (relative to the air) into two components, $v_n$ for normal velocity and $v_t$ for tangential velocity.

fig1

The two opposing forces act on that surface element $F_n$ being the pressure drag, and $F_t$ being the surface friction. They are not proportional to each other since the latter depends on air viscosity and the first on density.

Add up all the combined effects all around the body to get an idea of what the net forces and torques are.

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