Kinematics in multiple frames

[Bling Fizikst]

Homework Statement

A helicopter turns in a horizontal circle with $O$ moving in a circle of radius $R$ at speed $v$, which is increasing at the rate of $\dot{v}$ , $t=\frac{\pi}{4p}$ . The helicopter blades osscilate with $\theta=\theta_{\circ}+\theta_1 \sin pt$ . Find $\vec{v_{BO}} ,\vec{v_{AO}}$ at $t=\frac{\pi}{4p}$

Relevant Equations

below

We know $$\vec{v}_{B/O} \equiv \vec{v}_{B/1}$$ $$v_{O/F} = v\hat{e}_t$$ $$a_{O/F} = \dot{v}\hat{e}_t + \frac{v^2}{R} \hat{e}_n$$\$$\omega_{1/F} = -\omega \hat{j}$$ $$\omega_{2/1} = \dot{\theta} \hat{i}$$Using velocity transfer relations, $$v_{B/F} = v_{B/1} + v_{O/F} + \omega_{1/F} \times OB$$ $$v_{B/F} = v_{B/2} + v_{A/F} + \omega_{2/F} \times AB$$ $$v_{A/F} = v_{A/1} + v_{O/F} + \omega_{1/F} \times OA$$ By additivity of angular velocities, we can find $\omega_{2/F}$.Since $v_{B/2} = 0$and $v_{A/1} = 0$ , we get $$v_{B/1} = \hat{i} \left(-\omega a + \omega b \sin\theta - \omega \theta_1 b p \sin(pt) \right) + \hat{j} \left(\theta_1 b p \sin(pt) \sin\theta\right) + \hat{k} \left(\theta_1 b p \sin\theta \sin(pt)\right)$$

 

[kuruman]

Please edit your post to fix the LaTeX.

 

[Bling Fizikst]

I asked chatgpt to correct it yet it is not rendering here

 

[Orodruin]

I asked chatgpt to correct it yet it is not rendering here

You cannot use regular LaTeX delimiters. Writing LaTeX here requires different delimiters. See https://www.physicsforums.com/help/latexhelp/

 

[Bling Fizikst]

edited it , yet it is not rendering

 

[kuruman]

I asked chatgpt to correct it yet it is not rendering here

You cannot use regular LaTeX delimiters. Writing LaTeX here requires different delimiters. See https://www.physicsforums.com/help/latexhelp/

Or, even better, ask Chatgpt to follow the link that @Orodruin provided. It might "learn" something.

 

[erobz]

Homework Statement: A helicopter turns in a horizontal circle with $O$ moving in a circle of radius $R$ at speed $v$, which is increasing at the rate of $\dot{v}$ , $t=\frac{\pi}{4p}$ . The helicopter blades osscilate with $\theta=\theta_{\circ}+\theta_1 \sin pt$ . Find $\vec{v_{BO}} ,\vec{v_{AO}}$ at $t=\frac{\pi}{4p}$
Relevant Equations: below

View attachment 356710

We know $$\vec{v}_{B/O} \equiv \vec{v}_{B/1}$$ $$v_{O/F} = v\hat{e}_t$$ $$a_{O/F} = \dot{v}\hat{e}_t + \frac{v^2}{R} \hat{e}_n$$ $$\omega_{1/F} = -\omega \hat{j}$$ $$\omega_{2/1} = \dot{\theta} \hat{i}$$Using velocity transfer relations, $$v_{B/F} = v_{B/1} + v_{O/F} + \omega_{1/F} \times OB$$ $$v_{B/F} = v_{B/2} + v_{A/F} + \omega_{2/F} \times AB$$ $$v_{A/F} = v_{A/1} + v_{O/F} + \omega_{1/F} \times OA$$ By additivity of angular velocities, we can find $\omega_{2/F}$.Since $v_{B/2} = 0$and $v_{A/1} = 0$ , we get $$v_{B/1} = \hat{i} \left(-\omega a + \omega b \sin\theta - \omega \theta_1 b p \sin(pt) \right) + \hat{j} \left(\theta_1 b p \sin(pt) \sin\theta\right) + \hat{k} \left(\theta_1 b p \sin\theta \sin(pt)\right)$$

you had an extra "\" in there.

 

[Bling Fizikst]

you had an extra "\" in there.

thank you

 

[WWGD]

I asked chatgpt to correct it yet it is not rendering here

Use ## on both sides of the text to wrap the Latex.

 

[haruspex]

Homework Statement: A helicopter turns in a horizontal circle with $O$ moving in a circle of radius $R$

Not sure what that means. Is the helicopter moving in a circle about O, or does O represent the helicopter and that is moving in a circle about something else, or should it say that the helicopter blades move in a horizontal circle about O (which represents the circling helicopter)?

at speed $v$, which is increasing at the rate of $\dot{v}$ , $t=\frac{\pi}{4p}$ .

What does that equation for t mean there? Is it just (prematurely) stating the time at which $\theta$ is to be evaluated, or is it intended to allow us to evaluate $v$ similarly? If the latter, we would have to know $v(0)$.

 

[Lnewqban]

The helicopter blades oscillate with $\theta=\theta_{\circ}+\theta_1 \sin pt$ .

It seems that the cited oscillation is known as flapping of the helicopter blades.

Copied from
http://www.pilotfriend.com/training/flight_training/rotary/helis.htm

"Once the aircraft moves forward, it begins to change this balance. If we travel 10 MPH forward, then the forward moving, or advancing rotor blade, is experiencing 110 MPH of wind speed, and the rearward, or retreating blade, is experiencing only 90 MPH of wind speed.

When this happens, we get an unbalanced condition, and the advancing blade experiencing more lift wants to climb, while the retreating blade experiences less lift and wants to drop. This is where we get the term "Dis-Symmetry of lift". The lift is not symmetrical around the entire rotor system.

How do we compensate for this situation? We compensate by allowing the rotor to flap. By allowing the advancing blade to flap upward, and the retreating blade to flap downward, it changes the angle of incidence on both rotor blades which balances out the entire rotor system. As you can see in this simple graphic, there are a few ways to allow for blade flapping.

One is to allow the blades to flap on hinges (Articulated rotor system)."

​