Circular motion A-Level

  • #1

Homework Statement


An aircraft flies with its wings tilted, in order to fly in a horizontal circle of radius r. The aircraft has mass 4.00 x 104 Kg and has a constant speed of 250ms-1 . with the aircraft flying in this way, two forces are acting on the plane, P [lift] and W [weight which is given as 3.92*105 N] . P is acting at 35 degrees to the vertical. calculate P

Homework Equations


centripetal force=mass * velocity2 / radius
centripetal force= mass * angular speed2 * radius
angular speed = theta / time
angular speed = velocity/ radius
angular speed = 2pi * frequency
centripetal acceleration = velocity2 / radius
centripetal acceleration = angular speed2 * radius
Force = mass * acceleration

The Attempt at a Solution


I cant do it. I swear we need more information. can someone enlighten me.
 
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Answers and Replies

  • #2
PeroK
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Homework Statement


An aircraft flies with its wings tilted, in order to fly in a horizontal circle of radius r. The aircraft has mass 4.00 x 104 Kg and has a constant speed of 250ms-1 . with the aircraft flying in this way, two forces are acting on the plane, P [lift] and W [weight which is given as 3.92*103 N] . P is acting at 35 degrees to the vertical. calculate P

Homework Equations


centripetal force=mass * velocity2 / radius
centripetal force= mass * angular speed2 * radius
angular speed = theta / time
angular speed = velocity/ radius
angular speed = 2pi * frequency
centripetal acceleration = velocity2 / radius
centripetal acceleration = angular speed2 * radius

The Attempt at a Solution


I cant do it. I swear we need more information. can someone enlighten me.
What information do you think you need?

By the way, I think the weight should be ##3.92*10^5N##. Personally I would do this question algebraically and just plug the numbers in at the end. That's also a good way to work out what quantities you do need.
 
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  • #3
What information do you think you need?

By the way, I think the weight should be ##3.92*10^5N##. Personally I would do this question algebraically and just plug the numbers in at the end. That's also a good way to work out what quantities you do need.
yes I think that the weight is correct. I couldn't read the sheet correctly.
the info that I think we need : time or radius
I am not sure whether we need to plug it into an equation or just use trig??
 
  • #4
PeroK
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yes I think that the weight is correct. I couldn't read the sheet correctly.
the info that I think we need :
I am not sure whether we need to plug it into an equation or just use trig??
You need to think about the forces involved. There's a key equation that will open up the whole problem. And, by equation I mean two quantities that must be equal.

Hint: why doesn't the plan fall out of the sky?
 
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  • #5
You need to think about the forces involved. There's a key equation that will open up the whole problem. And, by equation I mean two quantities that must be equal.

Hint: why doesn't the plan fall out of the sky?
Correct me if I am wrong,
but can I say that the vertical component of P is PCos(35), and it doesn't fall out of the sky because Weight is equal to PCos(35)
Therefore, PCos(35) = 3.92 * 105
So, If we rearrange that equation, P = 3.92*105 / Cos(35) == 321107.6 [1dp]
??
 
  • #6
PeroK
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Correct me if I am wrong,
but can I say that the vertical component of P is PCos(35), and it doesn't fall out of the sky because Weight is equal to PCos(35)
Therefore, PCos(35) = 3.92 * 105
So, If we rearrange that equation, P = 3.92*105 / Cos(35)
??
Yes, that's it exactly. Although, I prefer:

##P = \frac{mg}{\cos \theta}##

or:

##P = \frac{W}{\cos \theta}##
 
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  • #7
Yes, that's it exactly. Although, I prefer:

##P = \frac{mg}{\cos \theta}##

or:

##P = \frac{W}{\cos \theta}##
Thanks,
Could you please explain to me how you came to those equations?
 
  • #8
PeroK
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Thanks,
Could you please explain to me how you came to those equations?
I just copied what you had done! I just used letters for the quantities involved instead of plugging in the numbers. There are a lot of advantages to this. Not least, once you plug in the numbers, you lose sight of what is related to what and if you make a mistake, well one number looks very like another, and it is hard to spot an error.

That said, I thought the question was to calculate ##r##. I see now that all you need is ##P##, which is a lot simpler.

Is part 2 of the question to calculate ##r##?
 
  • #9
I just copied what you had done! I just used letters for the quantities involved instead of plugging in the numbers. There are a lot of advantages to this. Not least, once you plug in the numbers, you lose sight of what is related to what and if you make a mistake, well one number looks very like another, and it is hard to spot an error.

That said, I thought the question was to calculate ##r##. I see now that all you need is ##P##, which is a lot simpler.

Is part 2 of the question to calculate ##r##?
Yes it is.

but I am pretty sure how to do it

F=mv2 / r ,, rearrange for r
 
  • #10
PeroK
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Yes it is.

but I am pretty sure how to do it

F=mv2 / r ,, rearrange for r
If you post your answer, I can show you the algebraic approach, which might be interesting and/or enlightening!
 

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