Angular velocity of an airplane

In summary: I like to see the derivation.For the case when ##t=3~\rm min##, the distance ##r## between you and the aircraft is the length of the hypotenuse. Note that you can't simply set ##v=900~\rm km/h## because ##v## in ##v=\omega r## is the magnitude of ##\vec v_\perp##, the component of the velocity perpendicular to ##\vec r##. That's what @jbriggs444 is getting at in post 13.The other approach to solving the problem is to forgo using ##v=\omega r## and use ##\omega = d\theta/dt
  • #1
24
2
Homework Statement
An aircraft passes directly over you with a speed of 900 km/h at an altitude of 10,000 m. What is the angular velocity of the aircraft (relative to you) when directly overhead? Three minutes later?
Relevant Equations:: Angular velocity
Relevant Equations
Angular velocity
Speed = 900km/hour
tan(α)=900t/10000 α=arctan(900t/10000)
Derivative is 900/(10000+81 t^2)
 
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  • #2
Zoubayr said:
Relevant Equations:: Angular velocity
That's not an equation. Equations have equal signs in them.

Zoubayr said:
Speed = 900km/hour
tan(α)=900t/10000 α=arctan(900t/10000)
Derivative is 900/(10000+81 t^2)
It's not clear what you're doing here. Please explain your reasoning.
 
  • #3
Zoubayr said:
Speed = 900km/hour
tan(α)=900t/10000 α=arctan(900t/10000)
Derivative is 900/(10000+81 t^2)
1. You have divided a speed in km/h by a distance in m. What are the units of the result?
2. You don't need to get into trig. You are only concerned with the instantaneous situation as the plane is directly overhead, so you can approximate for small t.
3. Never plug in numbers straight away. Work symbolically as far as you can. It has many advantages, including allowing others to follow what you are doing.
4. There is an easier way, which may become apparent if you resolve issues 2 and 3.
 
  • #4
vela said:
That's not an equation. Equations have equal signs in them.

The equation for angular velocity is v=rw

It's not clear what you're doing here. Please explain your reasoning.

Actually this is not my answer. A friend sent me this. I do not know how to start this question.
 
  • #5
Zoubayr said:
The equation for angular velocity is v=rw.
Well, that's a place to start. Can you identify any of the quantities given in the problem statement with the variables in that expression?
 
  • #6
From the eqn v=rw, only the value of v=900km/hr is given the question. Is the altitude considered as radius?
 
  • #7
Zoubayr said:
From the eqn v=rw, only the value of v=900km/hr is given the question. Is the altitude considered as radius?
In that equation, ##r## represents the distance from the point of reference to the moving object. What do you think? Does the altitude fit the bill?
 
  • #8
I think so but I am not sure
 
  • #9
Making a drawing can help.
drawing.jpg

What do you think now?
 
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  • #10
Yes the value of the altitude is the radius r. So by using v=rw and making w sof, we get: w=v/r= (900 x 1000)/ (10,000×3600) = 900,000/ 36,000,000 = 0.025 rad/s for t=0s. But what about when t=3s?
 
  • #11
Zoubayr said:
Yes the value of the altitude is the radius r. So by using v=rw and making w sof, we get: w=v/r= (900 x 1000)/ (10,000×3600) = 900,000/ 36,000,000 = 0.025 rad/s for t=0s. But what about when t=3s?
Try drawing a triangle where the angle ##\theta## subtends the horizontal distance traveled by the plane in time ##t##. What is the rate of change of that angle w.r.t. time?
 
  • #12
Zoubayr said:
Yes the value of the altitude is the radius r. So by using v=rw and making w sof, we get: w=v/r= (900 x 1000)/ (10,000×3600) = 900,000/ 36,000,000 = 0.025 rad/s for t=0s. But what about when t=3s?
Note that ##\tan\theta = \dfrac{vt}{r}##. What is the derivative of that with respect to time?
 
  • #13
erobz said:
Try drawing a triangle where the angle ##\theta## subtends the horizontal distance traveled by the plane in time ##t##. What is the rate of change of that angle w.r.t. time?
Or, if coordinates are more your thing, try splitting the velocity up into a radial component directly away from the ground station and a tangential component perpendicular to that.
 
  • #14
erobz said:
Try drawing a triangle where the angle ##\theta## subtends the horizontal distance traveled by the plane in time ##t##. What is the rate of change of that angle w.r.t. time?
CamScanner 12-04-2022 10.26.jpg

Is the drawing right?
 
  • #15
I would define as ##\theta## the other acute angle. See post #12. This problem is a one-liner if you answer my question in #12.
 
  • #16
Zoubayr said:
View attachment 318159
Is the drawing right?
I had the other angle in mind ( as @kuruman suggested). We should probably ask (based on what @jbriggs444 noted) are you supposed to be using calculus to solve the problem?
 
  • #17
erobz said:
I had the other angle in mind ( as @kuruman suggested). We should probably ask (based on what @jbriggs444 noted) are you supposed to be using calculus to solve the problem?
Even without calculus, one can ask "given a small increment of time, ##\Delta t##, how much does the angle change"? What is the ratio, ##\frac{\Delta \theta}{\Delta t}##?
 
  • #18
Without calculus, I would extract ##~\vec {\omega}~## from ##\vec v=\vec {\omega}\times \vec {r}~## by crossing on the left with ##\vec r.##
 
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  • #19
Zoubayr said:
Is the drawing right?
For the case when ##t=3~\rm min##, the distance ##r## between you and the aircraft is the length of the hypotenuse. Note that you can't simply set ##v=900~\rm km/h## because ##v## in ##v=\omega r## is the magnitude of ##\vec v_\perp##, the component of the velocity perpendicular to ##\vec r##. That's what @jbriggs444 is getting at in post 13.

The other approach to solving the problem is to forgo using ##v=\omega r## and use ##\omega = d\theta/dt## instead, where ##\theta## is the angle described by the others. You want to express the angle as a function of time and then differentiate.

I would suggest you solve the problem both ways. You should get the same answer either way.
 
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